翻译计划中，未完成
能力有限，如有错误请指出（其实是根本就不会翻译(╯‵□′)╯︵┻━┻）
玩弄位运算的黑客
由Sean Eron Anderson编写
由晨旭不负责任地胡乱翻译（powered by Google translate and Baidu translate）
特别说明，这些代码在任意场合 (除非另有说明) – 你可以随意使用它们。这些代码的收集由Sean Eron Anderson于1997-2005年完成。
这些代码和描述是为了让他们变得有用，但是我不做任何保证并且不会暗示保证这些代码可用。
截至2005年5月5日，所有的代码都经过了彻底的测试。已经有上千人阅读过这些代码。
此外，卡内基梅隆大学计算机系教授Randal Bryant，亲自用他的UCLID代码验证系统检验了几乎所有的程序。
他没测试过的代码也被我在32位机器上检测过了。对于第一个在代码中找到的bug，我将给予$10的报酬（通过支票或paypal）。如果对象是一个慈善机构，报酬将改为$20。
内容
Counting bits set
Counting bits set in 14, 24, or 32-bit words using 64-bit instructions
Count bits set (rank) from the most-significant bit upto a given position
Select the bit position (from the most-significant bit) with the given count (rank)
Computing parity (1 if an odd number of bits set, 0 otherwise)
Swapping Values
Reversing bit sequences
Reverse the bits in a byte with 3 operations (64-bit multiply and modulus division)
Reverse the bits in a byte with 4 operations (64-bit multiply, no division)
Reverse the bits in a byte with 7 operations (no 64-bit, only 32)
Reverse an N-bit quantity in parallel with 5 * lg(N) operations
Modulus division (aka computing remainders)
Computing modulus division by 1 << s without a division operation (obvious)
Computing modulus division by (1 << s) – 1 without a division operation
Computing modulus division by (1 << s) – 1 in parallel without a division operation
Finding integer log base 2 of an integer (aka the position of the highest bit set)
Find the log base 2 of an integer with the MSB N set in O(N) operations (the obvious way)
Find the integer log base 2 of an integer with an 64-bit IEEE float
Find the log base 2 of an N-bit integer in O(lg(N)) operations
Find the log base 2 of an N-bit integer in O(lg(N)) operations with multiply and lookup
Find integer log base 2 of the pow(2, r)-root of a 32-bit IEEE float (for unsigned integer r)
Counting consecutive trailing zero bits (or finding bit indices)
Count the consecutive zero bits (trailing) on the right linearly
Count the consecutive zero bits (trailing) on the right in parallel
Count the consecutive zero bits (trailing) on the right by binary search
Count the consecutive zero bits (trailing) on the right by casting to a float
Count the consecutive zero bits (trailing) on the right with modulus division and lookup
Count the consecutive zero bits (trailing) on the right with multiply and lookup
Interleaving bits (aka computing Morton Numbers)
Testing for ranges of bytes in a word (and counting occurances found)
关于运算次数的统计方法
这里运算次数的统计标准，任何一个操作符都被算作一次运算。
过程操作（？）没有被写进RAM所以不会被统计。
当然，这种计数方法得到的结果仅能作为实际CPU执行机器指令时间的参考。
假设所有运算都会使用相同的时间，这在现实中是不可能的，但是现在的处理器已经越来越向这个趋势发展了。（感觉现在应该差不多了吧2333 -译者吐槽道）
有很多细节都决定了系统将用什么速度来运行给定的代码，如缓存大小、内存带宽、指令集等。
最后，基准测试是测试一个算法是否比另一个快的最好方法，所以考虑下面的技术作为可能性，测试你的目标架构。（？）
判断整型的符号
int v; //我们想求出v的符号 int sign; //结果在这里 // CHAR_BIT是位/字节数（通常为8）。 sign = -(v < 0); // if v < 0 then -1, else 0. //或者，为了避免在CPU标志寄存器出现分支（英特尔32位体系架构） sign = -(int)((unsigned int)((int)v) >> (sizeof(int) * CHAR_BIT - 1));// 将其变成无符号数，逻辑右移，若为负，则为00000001；否则为00000000 //或者，对于一个较少指令（但不可移植）： sign = v >> (sizeof(int) * CHAR_BIT - 1);//数字右移，若为负，则 11111111 (-1值），否则00000000。
对于32位整数来说，上面的最后一个表达式的计算结果为 sign = v >> 31 for 32-bit integers。
这个算法比通常的算法相比要快, 通常是sign = -(v < 0).
这样算能得出结果的原因是，当无符号整型被右移时，最左边的字节将会被移到别的字节所在的位置。
当最左侧为1时，该值为0，否则为1。
然而，这个算法只适用于这种特殊结构。
此外，如果你更习惯用-1和1来表示结果，你可以用：
sign = +1 | (v >> (sizeof(int) * CHAR_BIT - 1)); // if v < 0 then -1, else +1
或者你如果倾向于用1、0和+1来表示结果，那么可以这样：
sign = (v != 0) | -(int)((unsigned int)((int)v) >> (sizeof(int) * CHAR_BIT - 1)); //或使用速度更快，但移植性更差的方法： sign = (v != 0) | (v >> (sizeof(int) * CHAR_BIT - 1)); // -1, 0, or +1 //或更简洁和可能更快的方法： sign = (v > 0) - (v < 0); // -1, 0, or +1
如果你想用0和1来表示是否为非负数，你可以用：
sign = 1 ^ ((unsigned int)v >> (sizeof(int) * CHAR_BIT - 1)); // if v < 0 then 0, else 1
（以下为机翻）
警告：2003年3月7日，安格斯杜根指出，1989年的ANSI C规范叶签署右移位实现定义的，因此在某些系统上这个技巧可能不工作的结果。
对于更大的便携性，托比斯佩特建议在2005 CHAR_BIT在此处使用9月28日，并在整个而不是假定字节是8位长。
安格斯建议更轻便版本以上，涉及浇铸3月4日，2006.Rohit加尔格建议版本为一个非负整数于2009年9月12日。
（原文）
Caveat: On March 7, 2003, Angus Duggan pointed out that the 1989 ANSI C
specification leaves the result of signed right-shift implementation-defined,
so on some systems this hack might not work. For greater portability,
Toby Speight suggested on September 28, 2005 that CHAR_BIT be used here
and throughout rather than assuming bytes were 8 bits long. Angus recommended
the more portable versions above, involving casting on March 4, 2006.Rohit Garg suggested the version
for non-negative integers on September 12, 2009.
检测两整数是否异号
int x, y; //输入数值进行比较 bool f = ((x ^ y) < 0); // x与y异号时为真
Manfred Weis在2009年11月26日建议我添加此内容。
不使用判断计算绝对值
int v; // 我们要计算v的绝对值 unsigned int r; //结果放于此变量内 int const mask = v >> sizeof(int) * CHAR_BIT - 1; r = (v + mask) ^ mask;
专利变化（？）Patented variation:
r = (v ^ mask) - mask;
一些CPU木有绝对值的指令集（或者编译器没有使用它们）。在有些使用判断语句缓慢的机器上，即使计算次数是相同的，这种方法也比普通方法计算更快r = (v < 0) ? -(unsigned)v : v
（出处及发展历程，不作翻译）
On March 7, 2003, Angus Duggan pointed out that the 1989 ANSI C specification leaves the result of signed right-shift implementation-defined,so on some systems this hack might not work. I've read that ANSI C does not require values to be represented as two's complement, so it may not work for that reason as well (on a diminishingly small number of old machines that still use one's complement).
On March 14, 2004, Keith H. Duggar sent me the patented variation above; it is superior to the one I initially came up with, r=(+1|(v>>(sizeof(int)*CHAR_BIT-1)))*v, because a multiply is not used.
Unfortunately, this method has been patented in the USA on June 6, 2000 by Vladimir Yu Volkonsky and assigned to Sun Microsystems.
On August 13, 2006, Yuriy Kaminskiy told me that the patent is likely invalid because the method was published well before the patent was even filed, such as in How to Optimize for the Pentium Processor by Agner Fog, dated November, 9, 1996. Yuriy also mentioned that this document was translated to Russian in 1997, which Vladimir could have read. Moreover, the Internet Archive also has an old link to it.
On January 30, 2007, Peter Kankowski shared with me an abs version he discovered that was inspired by Microsoft's Visual C++ compiler output. It is featured here as the primary solution.
On December 6, 2007, Hai Jin complained that the result was signed, so when computing the abs of the most negative value, it was still negative.
On April 15, 2008 Andrew Shapira pointed out that the obvious approach could overflow, as it lacked an (unsigned) cast then;for maximum portability he suggested(v < 0) ? (1 + ((unsigned)(-1-v))) : (unsigned)v.
But citing the ISO C99 spec on July 9, 2008, Vincent Lefèvre convinced me to remove it becasue even on non-2s-complement machines -(unsigned)v will do the right thing. The evaluation of -(unsigned)v first converts the negative value of v to an unsigned by adding 2**N, yielding a 2s complement representation of v's value that I'll call U. Then, U is negated, giving the desired result, -U = 0 – U = 2**N – U = 2**N – (v+2**N) = -v = abs(v).
不用判断求出两整型数中的最大或最小值
int x; //我们要求出x与y间的最小值 int y; int r; //结果放于这个变量内 r = y ^ ((x ^ y) & -(x < y)); // min(x, y)
在某些使用判断会非常耗时且不存在条件move指令（？）的机器上，用上面的方法速度会比较快，即使r = (x < y) ? x : y包括两个以上指令（但在通常情况下，这种显而易见的方法是最高效的）
代码的原理是（自己看去）:It works because if x<y,then -(x<y) will be all ones,so r= y ^ (x ^ y) & ~0 = y ^ x ^ y = x.Otherwise, if x>=y,then -(x<y) will be all zeros,so r= y ^ ((x ^ y) & 0) = y.On some machines, evaluating (x < y) as 0 or 1 requires a branch instruction, so there may be no advantage.
想寻找最大值，可以使用:
r = x ^ ((x ^ y) & -(x < y)); // max(x, y)
更快但是粗糙的版本:
如果你知道INT_MIN <= x – y <= INT_MAX，那么你可以使用下面的方法，能更快地运行且只需要运算一次。
r = y + ((x - y) & ((x - y) >> (sizeof(int) * CHAR_BIT - 1))); // min(x, y) r = x - ((x - y) & ((x - y) >> (sizeof(int) * CHAR_BIT - 1))); // max(x, y)
需要注意的是，1989的ANSI C规范并没有指定符号右移的结果，所以不是方便移植的方法。如果溢出时抛出异常，那么x和y的值可能是无符号整型。
（出处及发展历程，不作翻译）
On March 7, 2003, Angus Duggan pointed out the right-shift portability issue.On May 3, 2005, Randal E. Bryant alerted me to the need for the precondition, INT_MIN <= x-y <= INT_MAX,and suggested the non-quick and dirty version as a fix.
Both of these issues concern only the quick and dirty version.Nigel Horspoon observed on July 6, 2005 that gcc produced the same code on a Pentium as the obvious solution because of how it evaluates (x < y). On July 9, 2008 Vincent Lefèvre pointed out the potential for overflow exceptions with subtractions in r = y + ((x – y) & -(x < y)), which was the previous version.
Timothy B. Terriberry suggested using xor rather than add and subract to avoid casting and the risk of overflows on June 2, 2009.
判断一个整数是否是2的N次幂
unsigned int v; //我们想看看v是不是2的N次幂 bool f; //结果放于此变量 f = (v & (v - 1)) == 0;
请注意，在上述语句中，0将被错误的被认为是2的N次幂，可以用下面的语句解决：
f = v && !(v & (v - 1));
Sign extending from a constant bit-width
Sign extension is automatic for built-in types, such as chars and ints.
But suppose you have a signed two's complement number, x, that is stored
using only b bits. Moreover, suppose you want to convert x to an int,
which has more than b bits. A simple copy will work if x
is positive, but if negative, the sign must be extended. For example,
if we have only 4 bits to store a number, then -3 is represented as 1101
in binary. If we have 8 bits, then -3 is 11111101. The most-significant
bit of the 4-bit representation is replicated sinistrally to fill
in the destination when we convert to a representation with more bits;
this is sign extending.
In C, sign extension from a constant bit-width is trivial, since bit
fields may be specified in structs or unions.
For example, to convert from 5 bits to an full integer:
int x; // convert this from using 5 bits to a full int
int r; // resulting sign extended number goes here
struct {signed int x:5;} s;
r = s.x = x;The following is a C++ template function that uses the same
language feature to convert from B bits in one operation (though
the compiler is generating more, of course).
template <typename T, unsigned B>
inline T signextend(const T x)
{
struct {T x:B;} s;
return s.x = x;
}
int r = signextend<signed int,5>(x); // sign extend 5 bit number x to rJohn Byrd caught a typo in the code (attributed to html formatting)
on May 2, 2005. On March 4, 2006, Pat Wood pointed out that the ANSI C
standard requires that the bitfield have the keyword "signed" to be signed;
otherwise, the sign is undefined.
Sign extending from a variable bit-width
Sometimes we need to extend the sign of a number but we don't know a priori
the number of bits, b, in which it is represented. (Or we could be
programming in a language like Java, which lacks bitfields.)
unsigned b; // number of bits representing the number in x int x; // sign extend this b-bit number to r int r; // resulting sign-extended number int const m = 1U << (b - 1); // mask can be pre-computed if b is fixed x = x & ((1U << b) - 1); // (Skip this if bits in x above position b are already zero.) r = (x ^ m) - m;
The code above requires four operations, but when the bitwidth is a
constant rather than variable, it requires only two fast operations,
assuming the upper bits are already zeroes.
A slightly faster but less portable method that doesn't depend on
the bits in x above position b being zero is:
int const m = CHAR_BIT * sizeof(x) - b; r = (x << m) >> m;
Sean A. Irvine suggested that I
add sign extension methods to this page on June 13, 2004, and he providedm = (1 << (b - 1)) - 1; r = -(x & ~m) | x;as a starting point from which I optimized to get
m = 1U << (b – 1); r = -(x & m) | x.
But then on May 11, 2007, Shay Green suggested the version above,
which requires one less operation than mine. Vipin Sharma suggested
I add a step to deal with situations where x had possible ones in bits
other than the b bits we wanted to sign-extend on Oct. 15, 2008.
On December 31, 2009 Chris Pirazzi suggested I add the faster version,
which requires two operations for constant bit-widths and three
for variable widths.
Sign extending from a variable bit-width in 3 operations
The following may be slow on some machines, due to the effort required for
multiplication and division. This version is 4 operations. If you
know that your initial bit-width, b, is greater than 1, you might do this
type of sign extension in 3 operations by using
r= (x * multipliers[b]) / multipliers[b],
which requires only one array lookup.
unsigned b; // number of bits representing the number in x
int x; // sign extend this b-bit number to r
int r; // resulting sign-extended number
#define M(B) (1U << ((sizeof(x) * CHAR_BIT) - B)) // CHAR_BIT=bits/byte
static int const multipliers[] =
{
0, M(1), M(2), M(3), M(4), M(5), M(6), M(7),
M(8), M(9), M(10), M(11), M(12), M(13), M(14), M(15),
M(16), M(17), M(18), M(19), M(20), M(21), M(22), M(23),
M(24), M(25), M(26), M(27), M(28), M(29), M(30), M(31),
M(32)
}; // (add more if using more than 64 bits)
static int const divisors[] =
{
1, ~M(1), M(2), M(3), M(4), M(5), M(6), M(7),
M(8), M(9), M(10), M(11), M(12), M(13), M(14), M(15),
M(16), M(17), M(18), M(19), M(20), M(21), M(22), M(23),
M(24), M(25), M(26), M(27), M(28), M(29), M(30), M(31),
M(32)
}; // (add more for 64 bits)
#undef M
r = (x * multipliers[b]) / divisors[b];The following variation is not portable,
but on architectures that employ an arithmetic right-shift,
maintaining the sign, it should be fast.
const int s = -b; // OR: sizeof(x) * CHAR_BIT - b; r = (x << s) >> s;
Randal E. Bryant pointed out a bug on May 3, 2005 in an earlier version
(that used multipliers[] for divisors[]), where it failed on the case of
x=1 and b=1.
Conditionally set or clear bits without branching
bool f; // conditional flag unsigned int m; // the bit mask unsigned int w; // the word to modify: if (f) w |= m; else w &= ~m; w ^= (-f ^ w) & m; // OR, for superscalar CPUs: w = (w & ~m) | (-f & m);
On some architectures, the lack of branching can more than make up for
what appears to be twice as many operations. For instance, informal
speed tests on an AMD Athlon? XP 2100+ indicated it was 5-10%
faster. An Intel Core 2 Duo ran the superscalar version about 16%
faster than the first.
Glenn Slayden informed me of the first expression on
December 11, 2003. Marco Yu shared the superscalar version with me on
April 3, 2007 and alerted me to a typo 2 days later.
Conditionally negate a value without branching
If you need to negate only when a flag is false, then use the following
to avoid branching:
bool fDontNegate; // Flag indicating we should not negate v. int v; // Input value to negate if fDontNegate is false. int r; // result = fDontNegate ? v : -v; r = (fDontNegate ^ (fDontNegate - 1)) * v;
If you need to negate only when a flag is true, then use this:
bool fNegate; // Flag indicating if we should negate v. int v; // Input value to negate if fNegate is true. int r; // result = fNegate ? -v : v; r = (v ^ -fNegate) + fNegate;
Avraham Plotnitzky suggested I add the first version on June 2, 2009.
Motivated to avoid the multiply, I came up with the second version on
June 8, 2009. Alfonso De Gregorio pointed out that some parens were
missing on November 26, 2009, and received a bug bounty.
Merge bits from two values according to a mask
unsigned int a; // value to merge in non-masked bits unsigned int b; // value to merge in masked bits unsigned int mask; // 1 where bits from b should be selected; 0 where from a. unsigned int r; // result of (a & ~mask) | (b & mask) goes here r = a ^ ((a ^ b) & mask);
This shaves one operation from the obvious way of combining two sets of bits
according to a bit mask. If the mask is a constant, then there may be no
advantage.
Ron Jeffery sent this to me on February 9, 2006.
Counting bits set (naive way)
unsigned int v; // count the number of bits set in v
unsigned int c; // c accumulates the total bits set in v
for (c = 0; v; v >>= 1)
{
c += v & 1;
}The naive approach requires one iteration per bit, until no more
bits are set. So on a 32-bit word with only the high set,
it will go through 32 iterations.
Counting bits set by lookup table
static const unsigned char BitsSetTable256[256] =
{
# define B2(n) n, n+1, n+1, n+2
# define B4(n) B2(n), B2(n+1), B2(n+1), B2(n+2)
# define B6(n) B4(n), B4(n+1), B4(n+1), B4(n+2)
B6(0), B6(1), B6(1), B6(2)
};
unsigned int v; // count the number of bits set in 32-bit value v
unsigned int c; // c is the total bits set in v
// Option 1:
c = BitsSetTable256[v & 0xff] +
BitsSetTable256[(v >> 8) & 0xff] +
BitsSetTable256[(v >> 16) & 0xff] +
BitsSetTable256[v >> 24];
// Option 2:
unsigned char * p = (unsigned char *) &v;
c = BitsSetTable256[p[0]] +
BitsSetTable256[p[1]] +
BitsSetTable256[p[2]] +
BitsSetTable256[p[3]];
// To initially generate the table algorithmically:
BitsSetTable256[0] = 0;
for (int i = 0; i < 256; i++)
{
BitsSetTable256[i] = (i & 1) + BitsSetTable256[i / 2];
}On July 14, 2009 Hallvard Furuseth suggested the macro compacted table.
Counting bits set, Brian Kernighan's way
unsigned int v; // count the number of bits set in v
unsigned int c; // c accumulates the total bits set in v
for (c = 0; v; c++)
{
v &= v - 1; // clear the least significant bit set
}Brian Kernighan's method goes through as many iterations
as there are set bits. So if we have a 32-bit word with only
the high bit set, then it will only go once through the loop.
Published in 1988, the C Programming Language 2nd Ed.
(by Brian W. Kernighan and Dennis M. Ritchie)
mentions this in exercise 2-9.
On April 19, 2006 Don Knuth pointed out to me that this method
"was first published by Peter Wegner in CACM 3 (1960), 322.
(Also discovered independently by Derrick Lehmer and published
in 1964 in a book edited by Beckenbach.)"
Counting bits set in 14, 24, or 32-bit words using 64-bit instructions
unsigned int v; // count the number of bits set in v unsigned int c; // c accumulates the total bits set in v // option 1, for at most 14-bit values in v: c = (v * 0x200040008001ULL & 0x111111111111111ULL) % 0xf; // option 2, for at most 24-bit values in v: c = ((v & 0xfff) * 0x1001001001001ULL & 0x84210842108421ULL) % 0x1f; c += (((v & 0xfff000) >> 12) * 0x1001001001001ULL & 0x84210842108421ULL) % 0x1f; // option 3, for at most 32-bit values in v: c = ((v & 0xfff) * 0x1001001001001ULL & 0x84210842108421ULL) % 0x1f; c += (((v & 0xfff000) >> 12) * 0x1001001001001ULL & 0x84210842108421ULL) % 0x1f; c += ((v >> 24) * 0x1001001001001ULL & 0x84210842108421ULL) % 0x1f;
This method requires a 64-bit CPU with fast modulus division to be efficient.
The first option takes only 3 operations; the second option takes 10; and
the third option takes 15.
Rich Schroeppel originally created a 9-bit version, similiar to option 1;
see the Programming Hacks section ofBeeler, M., Gosper, R. W., and Schroeppel, R.
HAKMEM. MIT AI Memo 239, Feb. 29, 1972.His method was the inspiration for the variants above,
devised by Sean Anderson. Randal E. Bryant offered a couple bug fixes on
May 3, 2005. Bruce Dawson tweaked what had been a 12-bit version and made
it suitable for 14 bits using the same number of operations on
Feburary 1, 2007.
Counting bits set, in parallel
unsigned int v; // count bits set in this (32-bit value)
unsigned int c; // store the total here
static const int S[] = {1, 2, 4, 8, 16}; // Magic Binary Numbers
static const int B[] = {0x55555555, 0x33333333, 0x0F0F0F0F, 0x00FF00FF, 0x0000FFFF};
c = v - ((v >> 1) & B[0]);
c = ((c >> S[1]) & B[1]) + (c & B[1]);
c = ((c >> S[2]) + c) & B[2];
c = ((c >> S[3]) + c) & B[3];
c = ((c >> S[4]) + c) & B[4];The B array, expressed as binary, is:
B[0] = 0x55555555 = 01010101 01010101 01010101 01010101 B[1] = 0x33333333 = 00110011 00110011 00110011 00110011 B[2] = 0x0F0F0F0F = 00001111 00001111 00001111 00001111 B[3] = 0x00FF00FF = 00000000 11111111 00000000 11111111 B[4] = 0x0000FFFF = 00000000 00000000 11111111 11111111
We can adjust the method for larger integer sizes by continuing with
the patterns for the Binary Magic Numbers, B and S.
If there are k bits, then we need the arrays S and B to be ceil(lg(k))
elements long, and we must compute the same number of expressions for c as
S or B are long. For a 32-bit v, 16 operations are used.
The best method for counting bits in a 32-bit integer v is the following:
v = v - ((v >> 1) & 0x55555555); // reuse input as temporary v = (v & 0x33333333) + ((v >> 2) & 0x33333333); // temp c = ((v + (v >> 4) & 0xF0F0F0F) * 0x1010101) >> 24; // count
The best bit counting method takes only 12 operations,
which is the same as the lookup-table method,
but avoids the memory and potential cache misses of a table.
It is a hybrid between the purely parallel method above
and the earlier methods using multiplies (in the section on counting bits
with 64-bit instructions), though it doesn't use 64-bit instructions.
The counts of bits set in the bytes is done in parallel, and the sum
total of the bits set in the bytes is computed by multiplying by 0x1010101
and shifting right 24 bits.
A generalization of the best bit counting method to integers
of bit-widths upto 128 (parameterized by type T) is this:
v = v - ((v >> 1) & (T)~(T)0/3); // temp v = (v & (T)~(T)0/15*3) + ((v >> 2) & (T)~(T)0/15*3); // temp v = (v + (v >> 4)) & (T)~(T)0/255*15; // temp c = (T)(v * ((T)~(T)0/255)) >> (sizeof(T) - 1) * CHAR_BIT; // count
See Ian Ashdown's nice newsgroup post for more information
on counting the number of bits set (also known as sideways addition).
The best bit counting method was brought to my attention on October 5, 2005
by Andrew Shapira;
he found it in pages 187-188 ofSoftware
Optimization Guide for AMD Athlon? 64 and Opteron? Processors.
Charlie Gordon suggested a way to shave off one operation from the
purely parallel version on December 14, 2005, and Don Clugston trimmed three
more from it on December 30, 2005. I made a typo with Don's suggestion that
Eric Cole spotted on January 8, 2006. Eric later suggested the
arbitrary bit-width generalization to the best method on November 17, 2006.
On April 5, 2007, Al Williams observed that I had a line of dead code at the
top of the first method.
Count bits set (rank) from the most-significant bit upto a given position
The following finds the the rank of a bit, meaning it returns the sum of
bits that are set to 1 from the most-signficant bit downto the bit at
the given position.
uint64_t v; // Compute the rank (bits set) in v from the MSB to pos. unsigned int pos; // Bit position to count bits upto. uint64_t r; // Resulting rank of bit at pos goes here. // Shift out bits after given position. r = v >> (sizeof(v) * CHAR_BIT - pos); // Count set bits in parallel. // r = (r & 0x5555...) + ((r >> 1) & 0x5555...); r = r - ((r >> 1) & ~0UL/3); // r = (r & 0x3333...) + ((r >> 2) & 0x3333...); r = (r & ~0UL/5) + ((r >> 2) & ~0UL/5); // r = (r & 0x0f0f...) + ((r >> 4) & 0x0f0f...); r = (r + (r >> 4)) & ~0UL/17; // r = r % 255; r = (r * (~0UL/255)) >> ((sizeof(v) - 1) * CHAR_BIT);
Juha J?rvi sent this to me on November 21, 2009 as an inverse operation
to the computing the bit position with the given rank, which follows.
Select the bit position (from the most-significant bit) with the given count (rank)
The following 64-bit code selects the position of the rth 1 bit
when counting from the left. In other words if we start
at the most significant bit and proceed to the right,
counting the number of bits set to 1 until we reach the desired rank, r,
then the position where we stop is returned. If the rank requested exceeds
the count of bits set, then 64 is returned.
The code may be modified for 32-bit or counting from the right.
uint64_t v; // Input value to find position with rank r.
unsigned int r; // Input: bit's desired rank [1-64].
unsigned int s; // Output: Resulting position of bit with rank r [1-64]
uint64_t a, b, c, d; // Intermediate temporaries for bit count.
unsigned int t; // Bit count temporary.
// Do a normal parallel bit count for a 64-bit integer,
// but store all intermediate steps.
// a = (v & 0x5555...) + ((v >> 1) & 0x5555...);
a = v - ((v >> 1) & ~0UL/3);
// b = (a & 0x3333...) + ((a >> 2) & 0x3333...);
b = (a & ~0UL/5) + ((a >> 2) & ~0UL/5);
// c = (b & 0x0f0f...) + ((b >> 4) & 0x0f0f...);
c = (b + (b >> 4)) & ~0UL/0x11;
// d = (c & 0x00ff...) + ((c >> 8) & 0x00ff...);
d = (c + (c >> 8)) & ~0UL/0x101;
t = (d >> 32) + (d >> 48);
// Now do branchless select!
s = 64;
// if (r > t) {s -= 32; r -= t;}
s -= ((t - r) & 256) >> 3; r -= (t & ((t - r) >> 8));
t = (d >> (s - 16)) & 0xff;
// if (r > t) {s -= 16; r -= t;}
s -= ((t - r) & 256) >> 4; r -= (t & ((t - r) >> 8));
t = (c >> (s - 8)) & 0xf;
// if (r > t) {s -= 8; r -= t;}
s -= ((t - r) & 256) >> 5; r -= (t & ((t - r) >> 8));
t = (b >> (s - 4)) & 0x7;
// if (r > t) {s -= 4; r -= t;}
s -= ((t - r) & 256) >> 6; r -= (t & ((t - r) >> 8));
t = (a >> (s - 2)) & 0x3;
// if (r > t) {s -= 2; r -= t;}
s -= ((t - r) & 256) >> 7; r -= (t & ((t - r) >> 8));
t = (v >> (s - 1)) & 0x1;
// if (r > t) s--;
s -= ((t - r) & 256) >> 8;
s = 65 - s;If branching is fast on your target CPU, consider uncommenting the
if-statements and commenting the lines that follow them.
Juha J?rvi sent this to me on November 21, 2009.
Computing parity the naive way
unsigned int v; // word value to compute the parity of
bool parity = false; // parity will be the parity of v
while (v)
{
parity = !parity;
v = v & (v - 1);
}The above code uses an approach like Brian Kernigan's bit counting, above.
The time it takes is proportional to the number of bits set.
Compute parity by lookup table
static const bool ParityTable256[256] =
{
# define P2(n) n, n^1, n^1, n
# define P4(n) P2(n), P2(n^1), P2(n^1), P2(n)
# define P6(n) P4(n), P4(n^1), P4(n^1), P4(n)
P6(0), P6(1), P6(1), P6(0)
};
unsigned char b; // byte value to compute the parity of
bool parity = ParityTable256[b];
// OR, for 32-bit words:
unsigned int v;
v ^= v >> 16;
v ^= v >> 8;
bool parity = ParityTable256[v & 0xff];
// Variation:
unsigned char * p = (unsigned char *) &v;
parity = ParityTable256[p[0] ^ p[1] ^ p[2] ^ p[3]];Randal E. Bryant encouraged the addition of the (admittedly) obvious
last variation with variable p on May 3, 2005. Bruce Rawles found a typo
in an instance of the table variable's name on September 27, 2005, and
he received a $10 bug bounty. On October 9, 2006, Fabrice Bellard
suggested the 32-bit variations above, which require only one table lookup;
the previous version had four lookups (one per byte) and were slower.
On July 14, 2009 Hallvard Furuseth suggested the macro compacted table.
Compute parity of a byte using 64-bit multiply and modulus division
unsigned char b; // byte value to compute the parity of bool parity = (((b * 0x0101010101010101ULL) & 0x8040201008040201ULL) % 0x1FF) & 1;
The method above takes around 4 operations, but only works on bytes.
Compute parity of word with a multiply
The following method computes the parity of the 32-bit value in only 8
operations using a multiply.
unsigned int v; // 32-bit word v ^= v >> 1; v ^= v >> 2; v = (v & 0x11111111U) * 0x11111111U; return (v >> 28) & 1;
Also for 64-bits, 8 operations are still enough.
unsigned long long v; // 64-bit word v ^= v >> 1; v ^= v >> 2; v = (v & 0x1111111111111111UL) * 0x1111111111111111UL; return (v >> 60) & 1;
Andrew Shapira came up with this and sent it to me on Sept. 2, 2007.
Compute parity in parallel
unsigned int v; // word value to compute the parity of v ^= v >> 16; v ^= v >> 8; v ^= v >> 4; v &= 0xf; return (0x6996 >> v) & 1;
The method above takes around 9 operations, and works for 32-bit words.
It may be optimized to work just on bytes in 5 operations by removing the
two lines immediately following "unsigned int v;". The method first shifts
and XORs the eight nibbles of the 32-bit value together, leaving the result
in the lowest nibble of v. Next, the binary number 0110 1001 1001 0110
(0x6996 in hex) is shifted to the right by the value represented in the lowest
nibble of v. This number is like a miniature 16-bit parity-table indexed by
the low four bits in v. The result has the parity of v in bit 1, which is
masked and returned.
Thanks to Mathew Hendry for pointing out the shift-lookup idea at the end on
Dec. 15, 2002.
That optimization shaves two operations off using only shifting and XORing
to find the parity.
Swapping values with subtraction and addition
#define SWAP(a, b) ((&(a) == &(b)) || (((a) -= (b)), ((b) += (a)), ((a) = (b) - (a))))
This swaps the values of a and b without using a temporary variable.The initial check for a and b being the same location in memory may be
omitted when you know this can't happen. (The compiler may omit it anyway
as an optimization.) If you enable overflows exceptions, then pass unsigned
values so an exception isn't thrown.
The XOR method that follows may be slightly faster on some machines.
Don't use this with floating-point numbers (unless you operate on
their raw integer representations).
Sanjeev Sivasankaran suggested I add this on June 12, 2007.
Vincent Lefèvre pointed out the potential for overflow exceptions on
July 9, 2008
Swapping values with XOR
#define SWAP(a, b) (((a) ^= (b)), ((b) ^= (a)), ((a) ^= (b)))
This is an old trick to exchange the values of the variables a and bwithout using extra space for a temporary variable.
On January 20, 2005, Iain A. Fleming pointed out that the macro above
doesn't work when you swap with the same memory location, such as
SWAP(a[i], a[j]) with i == j. So if that may occur, consider
defining the macro as
(((a) == (b)) || (((a) ^= (b)), ((b) ^= (a)), ((a) ^= (b)))).
On July 14, 2009, Hallvard Furuseth suggested that on some machines,
(((a) ^ (b)) && ((b) ^= (a) ^= (b), (a) ^= (b))) might be faster,
since the (a) ^ (b) expression is reused.
Swapping individual bits with XOR
unsigned int i, j; // positions of bit sequences to swap unsigned int n; // number of consecutive bits in each sequence unsigned int b; // bits to swap reside in b unsigned int r; // bit-swapped result goes here unsigned int x = ((b >> i) ^ (b >> j)) & ((1U << n) - 1); // XOR temporary r = b ^ ((x << i) | (x << j));
As an example of swapping ranges of bits
suppose we have have b = 00101111
(expressed in binary) and we want to swap the n = 3 consecutive bits
starting at i = 1 (the second bit from the right) with the 3 consecutive
bits starting at j = 5; the result would be r =11100011 (binary).
This method of swapping is similar to the general purpose XOR swap
trick, but intended for operating on individual bits. The variable x
stores the result of XORing the pairs of bit values we want to swap,
and then the bits are set to the result of themselves XORed with x.
Of course, the result is undefined if the sequences overlap.
On July 14, 2009 Hallvard Furuseth suggested that I change the
1 << n to 1U << n
because the value was being assigned to an unsigned and to avoid shifting into
a sign bit.
Reverse bits the obvious way
unsigned int v; // input bits to be reversed
unsigned int r = v; // r will be reversed bits of v; first get LSB of v
int s = sizeof(v) * CHAR_BIT - 1; // extra shift needed at end
for (v >>= 1; v; v >>= 1)
{
r <<= 1;
r |= v & 1;
s--;
}
r <<= s; // shift when v's highest bits are zeroOn October 15, 2004, Michael Hoisie pointed out a bug in the original version.
Randal E. Bryant suggested removing an extra operation on May 3, 2005.
Behdad Esfabod suggested a slight change that eliminated one iteration of the
loop on May 18, 2005. Then, on February 6, 2007, Liyong Zhou suggested a
better version that loops while v is not 0, so rather than iterating over
all bits it stops early.
Reverse bits in word by lookup table
static const unsigned char BitReverseTable256[256] =
{
# define R2(n) n, n + 2*64, n + 1*64, n + 3*64
# define R4(n) R2(n), R2(n + 2*16), R2(n + 1*16), R2(n + 3*16)
# define R6(n) R4(n), R4(n + 2*4 ), R4(n + 1*4 ), R4(n + 3*4 )
R6(0), R6(2), R6(1), R6(3)
};
unsigned int v; // reverse 32-bit value, 8 bits at time
unsigned int c; // c will get v reversed
// Option 1:
c = (BitReverseTable256[v & 0xff] << 24) |
(BitReverseTable256[(v >> 8) & 0xff] << 16) |
(BitReverseTable256[(v >> 16) & 0xff] << 8) |
(BitReverseTable256[(v >> 24) & 0xff]);
// Option 2:
unsigned char * p = (unsigned char *) &v;
unsigned char * q = (unsigned char *) &c;
q[3] = BitReverseTable256[p[0]];
q[2] = BitReverseTable256[p[1]];
q[1] = BitReverseTable256[p[2]];
q[0] = BitReverseTable256[p[3]];The first method takes about 17 operations, and the second takes about 12,
assuming your CPU can load and store bytes easily.
On July 14, 2009 Hallvard Furuseth suggested the macro compacted table.
Reverse the bits in a byte with 3 operations (64-bit multiply and modulus division):
unsigned char b; // reverse this (8-bit) byte b = (b * 0x0202020202ULL & 0x010884422010ULL) % 1023;
The multiply operation creates five separate copies of the 8-bit
byte pattern to fan-out into a 64-bit value.
The AND operation selects the bits that are in the correct (reversed)
positions, relative to each 10-bit groups of bits.
The multiply and the AND operations copy the bits from the original
byte so they each appear in only one of the 10-bit sets.
The reversed positions of the bits from the original byte coincide with
their relative positions within any 10-bit set.
The last step, which involves modulus division by 2^10 – 1, has the
effect of merging together each set of 10 bits
(from positions 0-9, 10-19, 20-29, …) in the 64-bit value.
They do not overlap, so the addition steps underlying the
modulus division behave like or operations.
This method was attributed to Rich Schroeppel in the
Programming Hacks section ofBeeler, M., Gosper, R. W., and Schroeppel, R.
HAKMEM. MIT AI Memo 239, Feb. 29, 1972.
Reverse the bits in a byte with 4 operations (64-bit multiply, no division):
unsigned char b; // reverse this byte b = ((b * 0x80200802ULL) & 0x0884422110ULL) * 0x0101010101ULL >> 32;
The following shows the flow of the bit values with the boolean variablesa, b, c, d, e, f, g, and h, which
comprise an 8-bit byte. Notice how the first multiply fans out the
bit pattern to multiple copies, while the last multiply combines them
in the fifth byte from the right.
abcd efgh (-> hgfe dcba) * 1000 0000 0010 0000 0000 1000 0000 0010 (0x80200802) ------------------------------------------------------------------------------------------------- 0abc defg h00a bcde fgh0 0abc defg h00a bcde fgh0 & 0000 1000 1000 0100 0100 0010 0010 0001 0001 0000 (0x0884422110) ------------------------------------------------------------------------------------------------- 0000 d000 h000 0c00 0g00 00b0 00f0 000a 000e 0000 * 0000 0001 0000 0001 0000 0001 0000 0001 0000 0001 (0x0101010101) ------------------------------------------------------------------------------------------------- 0000 d000 h000 0c00 0g00 00b0 00f0 000a 000e 0000 0000 d000 h000 0c00 0g00 00b0 00f0 000a 000e 0000 0000 d000 h000 0c00 0g00 00b0 00f0 000a 000e 0000 0000 d000 h000 0c00 0g00 00b0 00f0 000a 000e 0000 0000 d000 h000 0c00 0g00 00b0 00f0 000a 000e 0000 ------------------------------------------------------------------------------------------------- 0000 d000 h000 dc00 hg00 dcb0 hgf0 dcba hgfe dcba hgfe 0cba 0gfe 00ba 00fe 000a 000e 0000 >> 32 ------------------------------------------------------------------------------------------------- 0000 d000 h000 dc00 hg00 dcb0 hgf0 dcba hgfe dcba & 1111 1111 ------------------------------------------------------------------------------------------------- hgfe dcba
Note that the last two steps can be combined on some processors because
the registers can be accessed as bytes;
just multiply so that a register stores the upper 32 bits of the result
and the take the low byte. Thus, it may take only 6 operations.
Devised by Sean Anderson, July 13, 2001.
Reverse the bits in a byte with 7 operations (no 64-bit):
b = ((b * 0x0802LU & 0x22110LU) | (b * 0x8020LU & 0x88440LU)) * 0x10101LU >> 16;
Make sure you assign or cast the result to an unsigned char to remove
garbage in the higher bits.
Devised by Sean Anderson, July 13, 2001. Typo spotted and correction supplied
by Mike Keith, January 3, 2002.
Reverse an N-bit quantity in parallel in 5 * lg(N) operations:
unsigned int v; // 32-bit word to reverse bit order // swap odd and even bits v = ((v >> 1) & 0x55555555) | ((v & 0x55555555) << 1); // swap consecutive pairs v = ((v >> 2) & 0x33333333) | ((v & 0x33333333) << 2); // swap nibbles ... v = ((v >> 4) & 0x0F0F0F0F) | ((v & 0x0F0F0F0F) << 4); // swap bytes v = ((v >> 8) & 0x00FF00FF) | ((v & 0x00FF00FF) << 8); // swap 2-byte long pairs v = ( v >> 16 ) | ( v << 16);
The following variation is also O(lg(N)), however it requires more
operations to reverse v. Its virtue is in taking less slightly memory
by computing the constants on the fly.
unsigned int s = sizeof(v) * CHAR_BIT; // bit size; must be power of 2
unsigned int mask = ~0;
while ((s >>= 1) > 0)
{
mask ^= (mask << s);
v = ((v >> s) & mask) | ((v << s) & ~mask);
}These methods above are best suited to situations where N is large.
If you use the above with 64-bit ints (or larger), then you need
to add more lines (following the pattern); otherwise only the lower
32 bits will be reversed and the result will be in the lower 32 bits.
See Dr. Dobb's Journal 1983, Edwin Freed's article on Binary Magic
Numbers for more information. The second variation was suggested
by Ken Raeburn on September 13, 2005. Veldmeijer mentioned that
the first version could do without ANDS in the last line
on March 19, 2006.
Compute modulus division by 1 << s without a division operator
const unsigned int n; // numerator const unsigned int s; const unsigned int d = 1U << s; // So d will be one of: 1, 2, 4, 8, 16, 32, ... unsigned int m; // m will be n % d m = n & (d - 1);
Most programmers learn this trick early, but it was included for the
sake of completeness.
Compute modulus division by (1 << s) – 1 without a division operator
unsigned int n; // numerator
const unsigned int s; // s > 0
const unsigned int d = (1 << s) - 1; // so d is either 1, 3, 7, 15, 31, ...).
unsigned int m; // n % d goes here.
for (m = n; n > d; n = m)
{
for (m = 0; n; n >>= s)
{
m += n & d;
}
}
// Now m is a value from 0 to d, but since with modulus division
// we want m to be 0 when it is d.
m = m == d ? 0 : m;This method of modulus division by an integer that is one less than a power
of 2 takes at most
5 + (4 + 5 * ceil(N / s)) * ceil(lg(N / s)) operations, where N is the
number of bits in the numerator. In other words, it takes at most
O(N * lg(N)) time.
Devised by Sean Anderson, August 15, 2001. Before Sean A. Irvine corrected me
on June 17, 2004, I mistakenly commented that we could alternatively assignm = ((m + 1) & d) - 1; at the end. Michael Miller spotted a
typo in the code April 25, 2005.
Compute modulus division by (1 << s) – 1 in parallel without a division
operator
// The following is for a word size of 32 bits!
static const unsigned int M[] =
{
0x00000000, 0x55555555, 0x33333333, 0xc71c71c7,
0x0f0f0f0f, 0xc1f07c1f, 0x3f03f03f, 0xf01fc07f,
0x00ff00ff, 0x07fc01ff, 0x3ff003ff, 0xffc007ff,
0xff000fff, 0xfc001fff, 0xf0003fff, 0xc0007fff,
0x0000ffff, 0x0001ffff, 0x0003ffff, 0x0007ffff,
0x000fffff, 0x001fffff, 0x003fffff, 0x007fffff,
0x00ffffff, 0x01ffffff, 0x03ffffff, 0x07ffffff,
0x0fffffff, 0x1fffffff, 0x3fffffff, 0x7fffffff
};
static const unsigned int Q[][6] =
{
{ 0, 0, 0, 0, 0, 0}, {16, 8, 4, 2, 1, 1}, {16, 8, 4, 2, 2, 2},
{15, 6, 3, 3, 3, 3}, {16, 8, 4, 4, 4, 4}, {15, 5, 5, 5, 5, 5},
{12, 6, 6, 6 , 6, 6}, {14, 7, 7, 7, 7, 7}, {16, 8, 8, 8, 8, 8},
{ 9, 9, 9, 9, 9, 9}, {10, 10, 10, 10, 10, 10}, {11, 11, 11, 11, 11, 11},
{12, 12, 12, 12, 12, 12}, {13, 13, 13, 13, 13, 13}, {14, 14, 14, 14, 14, 14},
{15, 15, 15, 15, 15, 15}, {16, 16, 16, 16, 16, 16}, {17, 17, 17, 17, 17, 17},
{18, 18, 18, 18, 18, 18}, {19, 19, 19, 19, 19, 19}, {20, 20, 20, 20, 20, 20},
{21, 21, 21, 21, 21, 21}, {22, 22, 22, 22, 22, 22}, {23, 23, 23, 23, 23, 23},
{24, 24, 24, 24, 24, 24}, {25, 25, 25, 25, 25, 25}, {26, 26, 26, 26, 26, 26},
{27, 27, 27, 27, 27, 27}, {28, 28, 28, 28, 28, 28}, {29, 29, 29, 29, 29, 29},
{30, 30, 30, 30, 30, 30}, {31, 31, 31, 31, 31, 31}
};
static const unsigned int R[][6] =
{
{0x00000000, 0x00000000, 0x00000000, 0x00000000, 0x00000000, 0x00000000},
{0x0000ffff, 0x000000ff, 0x0000000f, 0x00000003, 0x00000001, 0x00000001},
{0x0000ffff, 0x000000ff, 0x0000000f, 0x00000003, 0x00000003, 0x00000003},
{0x00007fff, 0x0000003f, 0x00000007, 0x00000007, 0x00000007, 0x00000007},
{0x0000ffff, 0x000000ff, 0x0000000f, 0x0000000f, 0x0000000f, 0x0000000f},
{0x00007fff, 0x0000001f, 0x0000001f, 0x0000001f, 0x0000001f, 0x0000001f},
{0x00000fff, 0x0000003f, 0x0000003f, 0x0000003f, 0x0000003f, 0x0000003f},
{0x00003fff, 0x0000007f, 0x0000007f, 0x0000007f, 0x0000007f, 0x0000007f},
{0x0000ffff, 0x000000ff, 0x000000ff, 0x000000ff, 0x000000ff, 0x000000ff},
{0x000001ff, 0x000001ff, 0x000001ff, 0x000001ff, 0x000001ff, 0x000001ff},
{0x000003ff, 0x000003ff, 0x000003ff, 0x000003ff, 0x000003ff, 0x000003ff},
{0x000007ff, 0x000007ff, 0x000007ff, 0x000007ff, 0x000007ff, 0x000007ff},
{0x00000fff, 0x00000fff, 0x00000fff, 0x00000fff, 0x00000fff, 0x00000fff},
{0x00001fff, 0x00001fff, 0x00001fff, 0x00001fff, 0x00001fff, 0x00001fff},
{0x00003fff, 0x00003fff, 0x00003fff, 0x00003fff, 0x00003fff, 0x00003fff},
{0x00007fff, 0x00007fff, 0x00007fff, 0x00007fff, 0x00007fff, 0x00007fff},
{0x0000ffff, 0x0000ffff, 0x0000ffff, 0x0000ffff, 0x0000ffff, 0x0000ffff},
{0x0001ffff, 0x0001ffff, 0x0001ffff, 0x0001ffff, 0x0001ffff, 0x0001ffff},
{0x0003ffff, 0x0003ffff, 0x0003ffff, 0x0003ffff, 0x0003ffff, 0x0003ffff},
{0x0007ffff, 0x0007ffff, 0x0007ffff, 0x0007ffff, 0x0007ffff, 0x0007ffff},
{0x000fffff, 0x000fffff, 0x000fffff, 0x000fffff, 0x000fffff, 0x000fffff},
{0x001fffff, 0x001fffff, 0x001fffff, 0x001fffff, 0x001fffff, 0x001fffff},
{0x003fffff, 0x003fffff, 0x003fffff, 0x003fffff, 0x003fffff, 0x003fffff},
{0x007fffff, 0x007fffff, 0x007fffff, 0x007fffff, 0x007fffff, 0x007fffff},
{0x00ffffff, 0x00ffffff, 0x00ffffff, 0x00ffffff, 0x00ffffff, 0x00ffffff},
{0x01ffffff, 0x01ffffff, 0x01ffffff, 0x01ffffff, 0x01ffffff, 0x01ffffff},
{0x03ffffff, 0x03ffffff, 0x03ffffff, 0x03ffffff, 0x03ffffff, 0x03ffffff},
{0x07ffffff, 0x07ffffff, 0x07ffffff, 0x07ffffff, 0x07ffffff, 0x07ffffff},
{0x0fffffff, 0x0fffffff, 0x0fffffff, 0x0fffffff, 0x0fffffff, 0x0fffffff},
{0x1fffffff, 0x1fffffff, 0x1fffffff, 0x1fffffff, 0x1fffffff, 0x1fffffff},
{0x3fffffff, 0x3fffffff, 0x3fffffff, 0x3fffffff, 0x3fffffff, 0x3fffffff},
{0x7fffffff, 0x7fffffff, 0x7fffffff, 0x7fffffff, 0x7fffffff, 0x7fffffff}
};
unsigned int n; // numerator
const unsigned int s; // s > 0
const unsigned int d = (1 << s) - 1; // so d is either 1, 3, 7, 15, 31, ...).
unsigned int m; // n % d goes here.
m = (n & M[s]) + ((n >> s) & M[s]);
for (const unsigned int * q = &Q[s][0], * r = &R[s][0]; m > d; q++, r++)
{
m = (m >> *q) + (m & *r);
}
m = m == d ? 0 : m; // OR, less portably: m = m & -((signed)(m - d) >> s);This method of finding modulus division by an integer that is one less
than a power of 2 takes at most O(lg(N)) time, where N is the number of
bits in the numerator (32 bits, for the code above).
The number of operations is at most 12 + 9 * ceil(lg(N)).
The tables may be removed if you know
the denominator at compile time; just extract the few relevent entries
and unroll the loop. It may be easily extended to more bits.
It finds the result by summing the values in base (1 << s) in parallel.
First every other base (1 << s) value is added to the previous one.
Imagine that the result is written on a piece of paper. Cut the paper
in half, so that half the values are on each cut piece. Align the values
and sum them onto a new piece of paper. Repeat by cutting this paper
in half (which will be a quarter of the size of the previous one)
and summing, until you cannot cut further. After performing lg(N/s/2)
cuts, we cut no more; just continue to add the values and put the result
onto a new piece of paper as before, while there are at least two s-bit
values.
Devised by Sean Anderson, August 20, 2001. A typo was spotted by
Randy E. Bryant on May 3, 2005 (after pasting the code, I had later
added "unsinged" to a variable declaration). As in the previous hack,
I mistakenly commented that we could alternatively assignm = ((m + 1) & d) - 1; at the end, and Don Knuth corrected
me on April 19, 2006 and suggestedm = m & -((signed)(m - d) >> s).
On June 18, 2009 Sean Irvine proposed a change that used((n >> s) & M[s]) instead of((n & ~M[s]) >> s),
which typically requires fewer operations because the M[s] constant is already
loaded.
Find the log base 2 of an integer with the MSB N set in O(N) operations
(the obvious way)
unsigned int v; // 32-bit word to find the log base 2 of
unsigned int r = 0; // r will be lg(v)
while (v >>= 1) // unroll for more speed...
{
r++;
}The log base 2 of an integer is the same as the position of the highest
bit set (or most significant bit set, MSB). The following log base 2
methods are faster than this one.
Find the integer log base 2 of an integer with an 64-bit IEEE float
int v; // 32-bit integer to find the log base 2 of
int r; // result of log_2(v) goes here
union { unsigned int u[2]; double d; } t; // temp
t.u[__FLOAT_WORD_ORDER==LITTLE_ENDIAN] = 0x43300000;
t.u[__FLOAT_WORD_ORDER!=LITTLE_ENDIAN] = v;
t.d -= 4503599627370496.0;
r = (t.u[__FLOAT_WORD_ORDER==LITTLE_ENDIAN] >> 20) - 0x3FF;The code above loads a 64-bit (IEEE-754 floating-point) double with
a 32-bit integer (with no paddding bits) by storing the integer in the
mantissa while the exponent is set to 252.
From this newly minted double,
252 (expressed as a double) is subtracted, which sets the
resulting exponent to the log base 2 of the input value, v. All that is
left is shifting the exponent bits into position (20 bits right) and
subtracting the bias, 0x3FF (which is 1023 decimal). This technique
only takes 5 operations, but many CPUs are slow at manipulating doubles,
and the endianess of the architecture must be accommodated.
Eric Cole sent me this on January 15, 2006. Evan Felix pointed out a typo
on April 4, 2006. Vincent Lefèvre told me on July 9, 2008 to
change the endian check to use the float's endian, which could differ
from the integer's endian.
Find the log base 2 of an integer with a lookup table
static const char LogTable256[256] =
{
#define LT(n) n, n, n, n, n, n, n, n, n, n, n, n, n, n, n, n
-1, 0, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3,
LT(4), LT(5), LT(5), LT(6), LT(6), LT(6), LT(6),
LT(7), LT(7), LT(7), LT(7), LT(7), LT(7), LT(7), LT(7)
};
unsigned int v; // 32-bit word to find the log of
unsigned r; // r will be lg(v)
register unsigned int t, tt; // temporaries
if (tt = v >> 16)
{
r = (t = tt >> 8) ? 24 + LogTable256[t] : 16 + LogTable256[tt];
}
else
{
r = (t = v >> 8) ? 8 + LogTable256[t] : LogTable256[v];
}The lookup table method takes only about 7 operations to find the log of
a 32-bit value. If extended for 64-bit quantities, it would take roughly
9 operations. Another operation can be trimmed off by using four tables,
with the possible additions incorporated into each. Using int table
elements may be faster, depending on your architecture.
The code above is tuned to uniformly distributed output values.
If your inputs are evenly distributed across all 32-bit values,
then consider using the following:
if (tt = v >> 24)
{
r = 24 + LogTable256[tt];
}
else if (tt = v >> 16)
{
r = 16 + LogTable256[tt];
}
else if (tt = v >> 8)
{
r = 8 + LogTable256[tt];
}
else
{
r = LogTable256[v];
}To initially generate the log table algorithmically:
LogTable256[0] = LogTable256[1] = 0;
for (int i = 2; i < 256; i++)
{
LogTable256[i] = 1 + LogTable256[i / 2];
}
LogTable256[0] = -1; // if you want log(0) to return -1Behdad Esfahbod and I shaved off a fraction of an operation (on average) on
May 18, 2005. Yet another fraction of an operation was removed on November
14, 2006 by Emanuel Hoogeveen. The variation that is tuned to evenly
distributed input values was suggested by David A. Butterfield on September
19, 2008. Venkat Reddy told me on January 5, 2009 that log(0) should return
-1 to indicate an error, so I changed the first entry in the table to that.
Find the log base 2 of an N-bit integer in O(lg(N)) operations
unsigned int v; // 32-bit value to find the log2 of
const unsigned int b[] = {0x2, 0xC, 0xF0, 0xFF00, 0xFFFF0000};
const unsigned int S[] = {1, 2, 4, 8, 16};
int i;
register unsigned int r = 0; // result of log2(v) will go here
for (i = 4; i >= 0; i--) // unroll for speed...
{
if (v & b[i])
{
v >>= S[i];
r |= S[i];
}
}
// OR (IF YOUR CPU BRANCHES SLOWLY):
unsigned int v; // 32-bit value to find the log2 of
register unsigned int r; // result of log2(v) will go here
register unsigned int shift;
r = (v > 0xFFFF) << 4; v >>= r;
shift = (v > 0xFF ) << 3; v >>= shift; r |= shift;
shift = (v > 0xF ) << 2; v >>= shift; r |= shift;
shift = (v > 0x3 ) << 1; v >>= shift; r |= shift;
r |= (v >> 1);
// OR (IF YOU KNOW v IS A POWER OF 2):
unsigned int v; // 32-bit value to find the log2 of
static const unsigned int b[] = {0xAAAAAAAA, 0xCCCCCCCC, 0xF0F0F0F0,
0xFF00FF00, 0xFFFF0000};
register unsigned int r = (v & b[0]) != 0;
for (i = 4; i > 0; i--) // unroll for speed...
{
r |= ((v & b[i]) != 0) << i;
}Of course, to extend the code to find the log of a 33- to 64-bit
number, we would append another element, 0xFFFFFFFF00000000, to b,
append 32 to S, and loop from 5 to 0. This method is much slower
than the earlier table-lookup version, but if you don't want big table
or your architecture is slow to access memory, it's a good choice.
The second variation involves slightly more operations, but it may be
faster on machines with high branch costs (e.g. PowerPC).
The second version was sent to me byEric Cole on January 7, 2006. Andrew Shapira subsequently trimmed a few operations
off of it and sent me his variation (above) on Sept. 1, 2007.
The third variation was suggested to me byJohn Owens on April 24, 2002; it's faster, butit is only suitable when the input is known to be a power of 2.
On May 25, 2003, Ken Raeburn suggested improving the general case by
using smaller numbers for b[], which load faster on some architectures
(for instance if the word size is 16 bits, then only one load instruction
may be needed). These values work for the general version, but not for
the special-case version below it, where v is a power of 2; Glenn Slayden
brought this oversight to my attention on December 12, 2003.
Find the log base 2 of an N-bit integer in O(lg(N)) operations with multiply and lookup
uint32_t v; // find the log base 2 of 32-bit v
int r; // result goes here
static const int MultiplyDeBruijnBitPosition[32] =
{
0, 9, 1, 10, 13, 21, 2, 29, 11, 14, 16, 18, 22, 25, 3, 30,
8, 12, 20, 28, 15, 17, 24, 7, 19, 27, 23, 6, 26, 5, 4, 31
};
v |= v >> 1; // first round down to one less than a power of 2
v |= v >> 2;
v |= v >> 4;
v |= v >> 8;
v |= v >> 16;
r = MultiplyDeBruijnBitPosition[(uint32_t)(v * 0x07C4ACDDU) >> 27];The code above computes the log base 2 of a 32-bit integer with
a small table lookup and multiply. It requires only 13 operations,
compared to (up to) 20 for the previous method. The purely table-based
method requires the fewest operations, but this offers a
reasonable compromise between table size and speed.
If you know that v is a power of 2, then you only need the following:
static const int MultiplyDeBruijnBitPosition2[32] =
{
0, 1, 28, 2, 29, 14, 24, 3, 30, 22, 20, 15, 25, 17, 4, 8,
31, 27, 13, 23, 21, 19, 16, 7, 26, 12, 18, 6, 11, 5, 10, 9
};
r = MultiplyDeBruijnBitPosition2[(uint32_t)(v * 0x077CB531U) >> 27];Eric Cole devised this January 8, 2006 after reading about the entry
below to round up to a power of 2 and
the method below forcomputing the number of trailing bits
with a multiply and lookup using a DeBruijn sequence.
On December 10, 2009, Mark Dickinson shaved off a couple operations
by requiring v be rounded up to one less than the next power of 2
rather than the power of 2.
Find integer log base 10 of an integer
unsigned int v; // non-zero 32-bit integer value to compute the log base 10 of
int r; // result goes here
int t; // temporary
static unsigned int const PowersOf10[] =
{1, 10, 100, 1000, 10000, 100000,
1000000, 10000000, 100000000, 1000000000};
t = (IntegerLogBase2(v) + 1) * 1233 >> 12; // (use a lg2 method from above)
r = t - (v < PowersOf10[t]);The integer log base 10 is computed by first using one of the techniques above
for finding the log base 2. By the relationship
log10(v) = log2(v) / log2(10), we need to
multiply it by 1/log2(10), which is approximately 1233/4096, or
1233 followed by a right shift of 12. Adding one is needed because the
IntegerLogBase2 rounds down. Finally, since the value t is only an
approximation that may be off by one, the exact value is found by
subtracting the result of v < PowersOf10[t].
This method takes 6 more operations than
IntegerLogBase2. It may be sped up (on machines with fast memory access)
by modifying the log base 2 table-lookup method above so that the entries
hold what is computed for t (that is, pre-add, -mulitply, and -shift).
Doing so would require a total of only 9 operations to find the log base 10,
assuming 4 tables were used (one for each byte of v).
Eric Cole suggested I add a version of this on January 7, 2006.
Find integer log base 10 of an integer the obvious way
unsigned int v; // non-zero 32-bit integer value to compute the log base 10 of int r; // result goes here r = (v >= 1000000000) ? 9 : (v >= 100000000) ? 8 : (v >= 10000000) ? 7 : (v >= 1000000) ? 6 : (v >= 100000) ? 5 : (v >= 10000) ? 4 : (v >= 1000) ? 3 : (v >= 100) ? 2 : (v >= 10) ? 1 : 0;
This method works well when the input is uniformly distributed over 32-bit
values because 76% of the inputs are caught by the first compare, 21%
are caught by the second compare, 2% are caught by the third, and so on
(chopping the remaining down by 90% with each comparision).
As a result, less than 2.6 operations are needed on average.
On April 18, 2007, Emanuel Hoogeveen suggested a variation on this where
the conditions used divisions, which were not as fast as simple comparisons.
Find integer log base 2 of a 32-bit IEEE float
const float v; // find int(log2(v)), where v > 0.0 && finite(v) && isnormal(v) int c; // 32-bit int c gets the result; c = *(const int *) &v; // OR, for portability: memcpy(&c, &v, sizeof c); c = (c >> 23) - 127;
The above is fast, but IEEE 754-compliant architectures utilizesubnormal (also called denormal)
floating point numbers. These have the exponent bits set to zero
(signifying pow(2,-127)),
and the mantissa is not normalized, so it contains leading zeros and
thus the log2 must be computed from the mantissa. To accomodate for
subnormal numbers, use the following:
const float v; // find int(log2(v)), where v > 0.0 && finite(v)
int c; // 32-bit int c gets the result;
int x = *(const int *) &v; // OR, for portability: memcpy(&x, &v, sizeof x);
c = x >> 23;
if (c)
{
c -= 127;
}
else
{ // subnormal, so recompute using mantissa: c = intlog2(x) - 149;
register unsigned int t; // temporary
// Note that LogTable256 was defined earlier
if (t = x >> 16)
{
c = LogTable256[t] - 133;
}
else
{
c = (t = x >> 8) ? LogTable256[t] - 141 : LogTable256[x] - 149;
}
}On June 20, 2004, Sean A. Irvine suggested that I include code to handle
subnormal numbers. On June 11, 2005, Falk Hüffner pointed out that
ISO C99 6.5/7 specified undefined behavior for the common type punning idiom
*(int *)&, though it has worked on 99.9% of C compilers.
He proposed using memcpy for maximum portability or a union with a float
and an int for better code generation than memcpy on some compilers.
Find integer log base 2 of the pow(2, r)-root of a 32-bit IEEE float
(for unsigned integer r)
const int r; const float v; // find int(log2(pow((double) v, 1. / pow(2, r)))), // where isnormal(v) and v > 0 int c; // 32-bit int c gets the result; c = *(const int *) &v; // OR, for portability: memcpy(&c, &v, sizeof c); c = ((((c - 0x3f800000) >> r) + 0x3f800000) >> 23) - 127;
So, if r is 0, for example, we have c = int(log2((double) v)).
If r is 1, then we have c = int(log2(sqrt((double) v))).
If r is 2, then we have c = int(log2(pow((double) v, 1./4))).
On June 11, 2005, Falk Hüffner pointed out that
ISO C99 6.5/7 left the type punning idiom *(int *)& undefined,
and he suggested using memcpy.
Count the consecutive zero bits (trailing) on the right linearly
unsigned int v; // input to count trailing zero bits
int c; // output: c will count v's trailing zero bits,
// so if v is 1101000 (base 2), then c will be 3
if (v)
{
v = (v ^ (v - 1)) >> 1; // Set v's trailing 0s to 1s and zero rest
for (c = 0; v; c++)
{
v >>= 1;
}
}
else
{
c = CHAR_BIT * sizeof(v);
}The average number of trailing zero bits in a (uniformly distributed)
random binary number is one, so this O(trailing zeros) solution isn't
that bad compared to the faster methods below.
Jim Cole suggested I add a linear-time method for counting the trailing zeros on August 15, 2007. On October 22, 2007, Jason Cunningham pointed out that
I had neglected to paste the unsigned modifier for v.
Count the consecutive zero bits (trailing) on the right in parallel
unsigned int v; // 32-bit word input to count zero bits on right unsigned int c = 32; // c will be the number of zero bits on the right v &= -signed(v); if (v) c--; if (v & 0x0000FFFF) c -= 16; if (v & 0x00FF00FF) c -= 8; if (v & 0x0F0F0F0F) c -= 4; if (v & 0x33333333) c -= 2; if (v & 0x55555555) c -= 1;
Here, we are basically doing the same operations as finding the log base 2
in parallel, but we first isolate the lowest 1 bit, and then proceed with
c starting at the maximum and decreasing.
The number of operations is at most 3 * lg(N) + 4, roughly, for N bit words.
Bill Burdick suggested an optimization, reducing the time from 4 * lg(N) on
February 4, 2011.
Count the consecutive zero bits (trailing) on the right by binary search
unsigned int v; // 32-bit word input to count zero bits on right
unsigned int c; // c will be the number of zero bits on the right,
// so if v is 1101000 (base 2), then c will be 3
// NOTE: if 0 == v, then c = 31.
if (v & 0x1)
{
// special case for odd v (assumed to happen half of the time)
c = 0;
}
else
{
c = 1;
if ((v & 0xffff) == 0)
{
v >>= 16;
c += 16;
}
if ((v & 0xff) == 0)
{
v >>= 8;
c += 8;
}
if ((v & 0xf) == 0)
{
v >>= 4;
c += 4;
}
if ((v & 0x3) == 0)
{
v >>= 2;
c += 2;
}
c -= v & 0x1;
}The code above is similar to the previous method, but it computes the number
of trailing zeros by accumulating c in a manner akin to binary search.
In the first step, it checks if the bottom 16 bits of v are zeros,
and if so, shifts v right 16 bits and adds 16 to c, which reduces the
number of bits in v to consider by half. Each of the subsequent
conditional steps likewise halves the number of bits until there is only 1.
This method is faster than the last one (by about 33%) because the bodies
of the if statements are executed less often.
Matt Whitlock suggested this on January 25, 2006. Andrew Shapira shaved
a couple operations off on Sept. 5, 2007 (by setting c=1 and unconditionally
subtracting at the end).
Count the consecutive zero bits (trailing)
on the right by casting to a float
unsigned int v; // find the number of trailing zeros in v int r; // the result goes here float f = (float)(v & -v); // cast the least significant bit in v to a float r = (*(uint32_t *)&f >> 23) - 0x7f;
Although this only takes about 6 operations, the time to convert
an integer to a float can be high on some machines.
The exponent of the 32-bit IEEE floating point
representation is shifted down, and the bias is subtracted to give
the position of the least significant 1 bit set in v.
If v is zero, then the result is -127.
Count the consecutive zero bits (trailing)
on the right with modulus division and lookup
unsigned int v; // find the number of trailing zeros in v
int r; // put the result in r
static const int Mod37BitPosition[] = // map a bit value mod 37 to its position
{
32, 0, 1, 26, 2, 23, 27, 0, 3, 16, 24, 30, 28, 11, 0, 13, 4,
7, 17, 0, 25, 22, 31, 15, 29, 10, 12, 6, 0, 21, 14, 9, 5,
20, 8, 19, 18
};
r = Mod37BitPosition[(-v & v) % 37];The code above finds the number of zeros that are trailing on the right, so
binary 0100 would produce 2. It makes use of the fact that the first 32
bit position values are relatively prime with 37, so performing a modulus
division with 37 gives a unique number from 0 to 36 for each. These numbers
may then be mapped to the number of zeros using a small lookup table.
It uses only 4 operations, however indexing into a table and performing modulus
division may make it unsuitable for some situations.
I came up with this independently and then searched for a subsequence of
the table values, and found it was invented earlier by Reiser, according toHacker's Delight.
Count the consecutive zero bits (trailing)
on the right with multiply and lookup
unsigned int v; // find the number of trailing zeros in 32-bit v
int r; // result goes here
static const int MultiplyDeBruijnBitPosition[32] =
{
0, 1, 28, 2, 29, 14, 24, 3, 30, 22, 20, 15, 25, 17, 4, 8,
31, 27, 13, 23, 21, 19, 16, 7, 26, 12, 18, 6, 11, 5, 10, 9
};
r = MultiplyDeBruijnBitPosition[((uint32_t)((v & -v) * 0x077CB531U)) >> 27];Converting bit vectors to indices of set bits is an example use for this.
It requires one more operation than the earlier one involving
modulus division, but the multiply may be faster. The expression
(v & -v) extracts the least significant 1 bit from v. The constant
0x077CB531UL is a de Bruijn sequence, which produces a unique pattern
of bits into the high 5 bits for each possible bit position that it is
multiplied against. When there are no bits set, it returns 0.
More information can be found by reading the paperUsing
de Bruijn Sequences to Index 1 in a Computer Word by Charles E.
Leiserson, Harald Prokof, and Keith H. Randall.
On October 8, 2005 Andrew Shapira suggested I add this. Dustin Spicuzza asked me on April 14, 2009 to
cast the result of the multiply to a 32-bit type so it would work when
compiled with 64-bit ints.
Round up to the next highest power of 2 by float casting
unsigned int const v; // Round this 32-bit value to the next highest power of 2
unsigned int r; // Put the result here. (So v=3 -> r=4; v=8 -> r=8)
if (v > 1)
{
float f = (float)v;
unsigned int const t = 1U << ((*(unsigned int *)&f >> 23) - 0x7f);
r = t << (t < v);
}
else
{
r = 1;
}The code above uses 8 operations, but works on all v <= (1<<31).
Quick and dirty version, for domain of 1 < v < (1<<25):
float f = (float)(v - 1); r = 1U << ((*(unsigned int*)(&f) >> 23) - 126);
Although the quick and dirty version only uses around 6 operations,
it is roughly three times slower than thetechnique below (which involves 12 operations) when
benchmarked on an Athlon? XP 2100+ CPU. Some CPUs will fare
better with it, though.
On September 27, 2005 Andi Smithers suggested I include a technique
for casting to floats to find the lg of a number for rounding up to
a power of 2. Similar to the quick and dirty version here,
his version worked with values less than (1<<25), due to mantissa
rounding, but it used one more operation.
Round up to the next highest power of 2
unsigned int v; // compute the next highest power of 2 of 32-bit v v--; v |= v >> 1; v |= v >> 2; v |= v >> 4; v |= v >> 8; v |= v >> 16; v++;
In 12 operations, this code computes the next highest power of 2 for a
32-bit integer.
The result may be expressed by the formula 1U << (lg(v – 1) + 1).
Note that in the edge case where v is 0, it returns 0, which isn't a
power of 2; you might append the expression v += (v == 0) to remedy this
if it matters.
It would be faster by 2 operations to use the formula and the
log base 2 method that uses a lookup table,
but in some situations, lookup tables are not suitable, so the above code
may be best. (On a Athlon? XP 2100+ I've found the above shift-left
and then OR code is as fast as using a single BSR assembly language
instruction, which scans in reverse to find the highest set bit.)
It works by copying the highest set bit to all of the lower
bits, and then adding one, which results in carries that set all of the lower
bits to 0 and one bit beyond the highest set bit to 1. If the original
number was a power of 2, then the decrement will reduce it to one less, so
that we round up to the same original value.
You might alternatively compute the next higher power of 2
in only 8 or 9 operations using a lookup table for
floor(lg(v)) and then evaluating 1<<(1+floor(lg(v)));
Atul Divekar suggested I mention this on September 5, 2010.
Devised by Sean Anderson, Sepember 14, 2001.
Pete Hart
pointed me toa
couple newsgroup posts by him and William Lewis in February of 1997,
where they arrive at the same algorithm.
Interleave bits the obvious way
unsigned short x; // Interleave bits of x and y, so that all of the
unsigned short y; // bits of x are in the even positions and y in the odd;
unsigned int z = 0; // z gets the resulting Morton Number.
for (int i = 0; i < sizeof(x) * CHAR_BIT; i++) // unroll for more speed...
{
z |= (x & 1U << i) << i | (y & 1U << i) << (i + 1);
}Interleaved bits (aka Morton numbers) are useful for linearizing 2D integer
coordinates, so x and y are combined into a single number that can be
compared easily and has the property that a number is usually close to
another if their x and y values are close.
Interleave bits by table lookup
static const unsigned short MortonTable256[256] =
{
0x0000, 0x0001, 0x0004, 0x0005, 0x0010, 0x0011, 0x0014, 0x0015,
0x0040, 0x0041, 0x0044, 0x0045, 0x0050, 0x0051, 0x0054, 0x0055,
0x0100, 0x0101, 0x0104, 0x0105, 0x0110, 0x0111, 0x0114, 0x0115,
0x0140, 0x0141, 0x0144, 0x0145, 0x0150, 0x0151, 0x0154, 0x0155,
0x0400, 0x0401, 0x0404, 0x0405, 0x0410, 0x0411, 0x0414, 0x0415,
0x0440, 0x0441, 0x0444, 0x0445, 0x0450, 0x0451, 0x0454, 0x0455,
0x0500, 0x0501, 0x0504, 0x0505, 0x0510, 0x0511, 0x0514, 0x0515,
0x0540, 0x0541, 0x0544, 0x0545, 0x0550, 0x0551, 0x0554, 0x0555,
0x1000, 0x1001, 0x1004, 0x1005, 0x1010, 0x1011, 0x1014, 0x1015,
0x1040, 0x1041, 0x1044, 0x1045, 0x1050, 0x1051, 0x1054, 0x1055,
0x1100, 0x1101, 0x1104, 0x1105, 0x1110, 0x1111, 0x1114, 0x1115,
0x1140, 0x1141, 0x1144, 0x1145, 0x1150, 0x1151, 0x1154, 0x1155,
0x1400, 0x1401, 0x1404, 0x1405, 0x1410, 0x1411, 0x1414, 0x1415,
0x1440, 0x1441, 0x1444, 0x1445, 0x1450, 0x1451, 0x1454, 0x1455,
0x1500, 0x1501, 0x1504, 0x1505, 0x1510, 0x1511, 0x1514, 0x1515,
0x1540, 0x1541, 0x1544, 0x1545, 0x1550, 0x1551, 0x1554, 0x1555,
0x4000, 0x4001, 0x4004, 0x4005, 0x4010, 0x4011, 0x4014, 0x4015,
0x4040, 0x4041, 0x4044, 0x4045, 0x4050, 0x4051, 0x4054, 0x4055,
0x4100, 0x4101, 0x4104, 0x4105, 0x4110, 0x4111, 0x4114, 0x4115,
0x4140, 0x4141, 0x4144, 0x4145, 0x4150, 0x4151, 0x4154, 0x4155,
0x4400, 0x4401, 0x4404, 0x4405, 0x4410, 0x4411, 0x4414, 0x4415,
0x4440, 0x4441, 0x4444, 0x4445, 0x4450, 0x4451, 0x4454, 0x4455,
0x4500, 0x4501, 0x4504, 0x4505, 0x4510, 0x4511, 0x4514, 0x4515,
0x4540, 0x4541, 0x4544, 0x4545, 0x4550, 0x4551, 0x4554, 0x4555,
0x5000, 0x5001, 0x5004, 0x5005, 0x5010, 0x5011, 0x5014, 0x5015,
0x5040, 0x5041, 0x5044, 0x5045, 0x5050, 0x5051, 0x5054, 0x5055,
0x5100, 0x5101, 0x5104, 0x5105, 0x5110, 0x5111, 0x5114, 0x5115,
0x5140, 0x5141, 0x5144, 0x5145, 0x5150, 0x5151, 0x5154, 0x5155,
0x5400, 0x5401, 0x5404, 0x5405, 0x5410, 0x5411, 0x5414, 0x5415,
0x5440, 0x5441, 0x5444, 0x5445, 0x5450, 0x5451, 0x5454, 0x5455,
0x5500, 0x5501, 0x5504, 0x5505, 0x5510, 0x5511, 0x5514, 0x5515,
0x5540, 0x5541, 0x5544, 0x5545, 0x5550, 0x5551, 0x5554, 0x5555
};
unsigned short x; // Interleave bits of x and y, so that all of the
unsigned short y; // bits of x are in the even positions and y in the odd;
unsigned int z; // z gets the resulting 32-bit Morton Number.
z = MortonTable256[y >> 8] << 17 |
MortonTable256[x >> 8] << 16 |
MortonTable256[y & 0xFF] << 1 |
MortonTable256[x & 0xFF];For more speed, use an additional table with values that are
MortonTable256 pre-shifted one bit to the left. This second table
could then be used for the y lookups, thus reducing the
operations by two, but almost doubling the memory required.
Extending this same idea, four tables could be used, with two of them
pre-shifted by 16 to the left of the previous two, so that we would
only need 11 operations total.
Interleave bits with 64-bit multiply
In 11 operations, this version interleaves bits of two bytes
(rather than shorts, as in the other versions),
but many of the operations are 64-bit multiplies
so it isn't appropriate for all machines. The input parameters, x and y,
should be less than 256.
unsigned char x; // Interleave bits of (8-bit) x and y, so that all of the unsigned char y; // bits of x are in the even positions and y in the odd; unsigned short z; // z gets the resulting 16-bit Morton Number. z = ((x * 0x0101010101010101ULL & 0x8040201008040201ULL) * 0x0102040810204081ULL >> 49) & 0x5555 | ((y * 0x0101010101010101ULL & 0x8040201008040201ULL) * 0x0102040810204081ULL >> 48) & 0xAAAA;
Holger Bettag was inspired to suggest this technique on
October 10, 2004 after reading the multiply-based bit reversals here.
Interleave bits by Binary Magic Numbers
static const unsigned int B[] = {0x55555555, 0x33333333, 0x0F0F0F0F, 0x00FF00FF};
static const unsigned int S[] = {1, 2, 4, 8};
unsigned int x; // Interleave lower 16 bits of x and y, so the bits of x
unsigned int y; // are in the even positions and bits from y in the odd;
unsigned int z; // z gets the resulting 32-bit Morton Number.
// x and y must initially be less than 65536.
x = (x | (x << S[3])) & B[3];
x = (x | (x << S[2])) & B[2];
x = (x | (x << S[1])) & B[1];
x = (x | (x << S[0])) & B[0];
y = (y | (y << S[3])) & B[3];
y = (y | (y << S[2])) & B[2];
y = (y | (y << S[1])) & B[1];
y = (y | (y << S[0])) & B[0];
z = x | (y << 1);Determine if a word has a zero byte
// Fewer operations: unsigned int v; // 32-bit word to check if any 8-bit byte in it is 0 bool hasZeroByte = ~((((v & 0x7F7F7F7F) + 0x7F7F7F7F) | v) | 0x7F7F7F7F);
The code above may be useful when doing a fast string copy in which a word
is copied at a time; it uses 5 operations.
On the other hand, testing for a null byte in the obvious ways (which follow)
have at least 7 operations (when counted in the most sparing way), and at most 12.
// More operations: bool hasNoZeroByte = ((v & 0xff) && (v & 0xff00) && (v & 0xff0000) && (v & 0xff000000)) // OR: unsigned char * p = (unsigned char *) &v; bool hasNoZeroByte = *p && *(p + 1) && *(p + 2) && *(p + 3);
The code at the beginning of this section (labeled "Fewer operations")
works by first zeroing the high bits of the 4 bytes in the word.
Subsequently, it adds a number that will result in an overflow to
the high bit of a byte if any of the low bits were initialy set.
Next the high bits of the original word are ORed with these values;
thus, the high bit of a byte is set iff any bit in the byte was set.
Finally, we determine if any of these high bits are zero by ORing with
ones everywhere except the high bits and inverting the result.
Extending to 64 bits is trivial; simply increase the constants to be
0x7F7F7F7F7F7F7F7F.
For an additional improvement, a fast pretest that requires only 4 operations
may be performed to determine if the word may have a zero byte.
The test also returns true if the high byte is 0x80, so there are
occasional false positives, but the slower and more reliable version
above may then be used on candidates for an overall increase in speed with
correct output.
bool hasZeroByte = ((v + 0x7efefeff) ^ ~v) & 0x81010100;
if (hasZeroByte) // or may just have 0x80 in the high byte
{
hasZeroByte = ~((((v & 0x7F7F7F7F) + 0x7F7F7F7F) | v) | 0x7F7F7F7F);
}There is yet a faster method —
use hasless(v, 1),
which is defined below; it
works in 4 operations and requires no subsquent verification. It simplifies
to
#define haszero(v) (((v) - 0x01010101UL) & ~(v) & 0x80808080UL)
The subexpression (v – 0x01010101UL), evaluates to a high bit set in any
byte whenever the corresponding byte in v is zero or greater than 0x80.
The sub-expression ~v & 0x80808080UL
evaluates to high bits set in bytes where the byte of v doesn't have its high
bit set (so the byte was less than 0x80). Finally, by ANDing these two
sub-expressions the result is the high bits set where the bytes in v
were zero, since the high bits set due to a value greater than 0x80
in the first sub-expression are masked off by the second.
Paul Messmer suggested the fast pretest improvement on October 2, 2004.
Juha J?rvi later suggested hasless(v, 1)on April 6, 2005, which
he found on Paul
Hsieh's Assembly Lab; previously it was written in a newsgroup post
on April 27, 1987 by Alan Mycroft.
Determine if a word has a byte equal to n
We may want to know if any byte in a word has a specific value. To do so,
we can XOR the value to test with a word that has been filled with the
byte values in which we're interested. Because XORing a value with itself
results in a zero byte and nonzero otherwise, we can pass the result tohaszero.
#define hasvalue(x,n) (haszero((x) ^ (~0UL/255 * (n))))
Stephen M Bennet suggested this on December 13, 2009 after reading the entry
for haszero.
Determine if a word has a byte less than n
Test if a word x contains an unsigned byte with value < n.
Specifically for n=1, it can be used to find a 0-byte by examining one
long at a time, or any byte by XORing x with a mask first.
Uses 4 arithmetic/logical operations when n is constant.
Requirements: x>=0; 0<=n<=128
#define hasless(x,n) (((x)-~0UL/255*(n))&~(x)&~0UL/255*128)
To count the number of bytes in x that are less than n in 7 operations, use
#define countless(x,n) (((~0UL/255*(127+(n))-((x)&~0UL/255*127))&~(x)&~0UL/255*128)/128%255)
Juha J?rvi sent this clever technique to me on April 6, 2005. Thecountless macro was added by Sean Anderson on
April 10, 2005, inspired by Juha's countmore, below.
Determine if a word has a byte greater than n
Test if a word x contains an unsigned byte with value > n.
Uses 3 arithmetic/logical operations when n is constant.
Requirements: x>=0; 0<=n<=127
#define hasmore(x,n) (((x)+~0UL/255*(127-(n))|(x))&~0UL/255*128)
To count the number of bytes in x that are more than n in 6 operations, use:
#define countmore(x,n) (((((x)&~0UL/255*127)+~0UL/255*(127-(n))|(x))&~0UL/255*128)/128%255)
The macro hasmore was suggested by Juha J?rvi on
April 6, 2005, and he added countmore on April 8, 2005.
Determine if a word has a byte between m and n
When m<n, this technique tests if a word x contains an
unsigned byte value, such that m < value < n.
It uses 7 arithmetic/logical operations when n and m are constant.
Note: Bytes that equal n can be reported by likelyhasbetweenas false positives,
so this should be checked by character if a certain result is needed.
Requirements: x>=0; 0<=m<=127; 0<=n<=128
#define likelyhasbetween(x,m,n) ((((x)-~0UL/255*(n))&~(x)&((x)&~0UL/255*127)+~0UL/255*(127-(m)))&~0UL/255*128)
This technique would be suitable for a fast pretest. A variation that
takes one more operation (8 total for constant m and n)
but provides the exact answer is:
#define hasbetween(x,m,n) ((~0UL/255*(127+(n))-((x)&~0UL/255*127)&~(x)&((x)&~0UL/255*127)+~0UL/255*(127-(m)))&~0UL/255*128)
To count the number of bytes in x that are between m and n (exclusive)
in 10 operations, use:
#define countbetween(x,m,n) (hasbetween(x,m,n)/128%255)
Juha J?rvi suggested likelyhasbetween on April 6, 2005.
From there,
Sean Anderson created hasbetween andcountbetween on April 10, 2005.
Compute the lexicographically next bit permutation
Suppose we have a pattern of N bits set to 1 in an integer and we want the
next permutation of N 1 bits in a lexicographical sense.
For example, if N is 3 and the bit pattern is 00010011, the next patterns
would be 00010101, 00010110, 00011001,00011010, 00011100, 00100011,
and so forth. The following is a fast way to compute the next permutation.
unsigned int v; // current permutation of bits unsigned int w; // next permutation of bits unsigned int t = v | (v - 1); // t gets v's least significant 0 bits set to 1 // Next set to 1 the most significant bit to change, // set to 0 the least significant ones, and add the necessary 1 bits. w = (t + 1) | (((~t & -~t) - 1) >> (__builtin_ctz(v) + 1));
The __builtin_ctz(v) GNU C compiler intrinsic for x86 CPUs returns the
number of trailing zeros. If you are using Microsoft compilers for x86,
the intrinsic is _BitScanForward. These both emit a bsf instruction, but
equivalents may be available for other architectures. If not, then
consider using one of the methods for counting the consecutive zero bits
mentioned earlier.
Here is another version that tends to be slower because of its
division operator, but it does not require counting the trailing zeros.
unsigned int t = (v | (v - 1)) + 1; w = t | ((((t & -t) / (v & -v)) >> 1) - 1);
Thanks to Dario Sneidermanis of Argentina, who provided this on
November 28, 2009.
A Belorussian translation (provided by Webhostingrating)
is available.
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这次是真的长，太长了