可能重复:
如何计算32位整数中的设置位数?
给出一个unsigned char类型值,计算它中的总位数.最快的方法是什么?我写了三个函数如下,最好的方法是什么,有人能想出一个更快的吗?(我只想要极快的一个)
const int tbl[] =
{
#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)
};
char naivecount (unsigned char val)
{
char cnt = 0;
while (val)
{
cnt += (val & 1);
val = val >> 1;
}
return cnt;
}
inline tableLookUp(int val)
{
assert(val >= 0 && val <= 255);
return tbl[val];
}
int asmCount(int val)
{
int res = 0;
asm volatile("xor %0, %0\n\t"
"begin:\n\t"
"cmp $0x0, %1\n\t"
"jle end\n\t"
"movl %1, %%ecx\n\t"
"and $0x1, %%ecx\n\t"
"addl %%ecx, %0\n\t"
"shrl %1\n\t"
"jmp begin\n\t"
"end:"
: "=r"(res)
: "r" (val));
return res;
}
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我测试了所有的方法,最快的就是使用 popcntl指令.在没有指令的平台上,我将使用表查找.
如果您想手动编码,请尝试以下方法:
#include <stdint.h>
int popcnt8(uint8_t x) {
x = (x & 0x55) + (x >> 1 & 0x55);
x = (x & 0x33) + (x >> 2 & 0x33);
x = (x & 0x0f) + (x >> 4 & 0x0f);
return x;
}
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在x86上,这编译为(AT&T-syntax):
popcnt8:
movl %edi, %eax
shrb %dil
andl $85, %eax
andl $85, %edi
addl %eax, %edi
movl %edi, %eax
shrb $2, %dil
andl $51, %eax
andl $51, %edi
addl %eax, %edi
movl %edi, %eax
shrb $4, %dil
andl $15, %eax
addl %edi, %eax
movzbl %al, %eax
ret
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将此与gcc与内在函数生成的内容进行比较:
#include <stdint.h>
int popcnt8_intrin(uint8_t x) { return __builtin_popcount(x); }
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在带有SSE 4.2的x86上:
popcnt8_intrin:
movzbl %dil, %eax
popcntl %eax, %eax
ret
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这不是最佳的; clang生成:
popcnt8_intrin:
popcntl %edi,%eax
ret
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将计算减少到一个(!)指令.
在没有SSE 4.2的x86上:
popcnt8_intrin:
subq $8, %rsp
movzbl %dil, %edi
call __popcountdi2
addq $8, %rsp
ret
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gcc基本上在这里调用它的库.不太理想.clang做得好一点:
popcnt8_intrin: # @popcnt8_intrin
movl %edi, %eax
shrl %eax
andl $85, %eax
subl %eax, %edi
movl %edi, %eax
andl $858993459, %eax # imm = 0x33333333
shrl $2, %edi
andl $858993459, %edi # imm = 0x33333333
addl %eax, %edi
movl %edi, %eax
shrl $4, %eax
addl %edi, %eax
andl $252645135, %eax # imm = 0xF0F0F0F
imull $16843009, %eax, %eax # imm = 0x1010101
shrl $24, %eax
ret
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clang计算整个32位数的popcnt.这不是最佳的imho.