我是汇编语言的新手，我有以下问题：
int main(){
int x = 2023; //nine 1 and two 0 as in bin it's - 11111100111
int y;
asm(
"mov eax, %1 ;"
"mov ecx, 32 ;"
"mov ebx, 0 ;"
"loop:"
"add eax, eax ;"
"jnc next ;"
"inc ebx ;"
"next :"
"dec ecx ;"
"jnz loop;"
"mov %0, ebx ;"
: "=r" (y)
: "r"(x)
: "eax", "ebx", "ecx"
);
代码编译使用
-masm=intel我想让这个程序计算 x 的二进制表示中 0 的个数。现在它可以在其中计数 1 但我无法弄清楚需要更改的内容。我试图改变几个功能，但它要么给我错误的价值，要么价值相同。
并澄清给定的代码。现在它给出的结果很好，因为它在 2023 年计数为 1，但我想以某种方式反转它以使其计数为 0 而不是 1。我通过更改
addincdec一个与您已经编写的代码保持接近的解决方案。 （显然，您遵循关于“使用 add / jnc 循环”的约束。）
; "mov eax, %1 ;"
mov ebx, 0
mov ecx, 32
L1: add eax, eax
jc L3 ; Found highest 1 bit
dec ecx
jnz L1
jmp L4 ; EAX was 0
L2: add eax, eax
jc L3
inc ebx ; Count zeroes
L3: dec ecx
jnz L2
L4: ; "mov %0, ebx ;"
unsigned int count_nonleading_zero_bits( unsigned int n ) {
unsigned int count = 0;
while ( n ) {
if ( !( n & 1 ) )
++count;
n >>= 1;
}
return count;
}
https://godbolt.org/z/asds67nbr
count_nonleading_zero_bits:
xor eax, eax
test edi, edi
je .L5
.L4:
mov edx, edi
and edx, 1
cmp edx, 1
adc eax, 0
shr edi
jne .L4
.L5:
ret
对于带有 BMI1 的现代 x86，您需要使用
32-popcnt(x)lzcnt(x)在 GNU C 中，看起来像
int count_significant_zeros(unsigned x)
{
if (!x)
return 0; // __builtin_clz has an undefined result for input 0
return sizeof(x)*CHAR_BIT - __builtin_popcount(x) - __builtin_clz(x);
// assume no padding bits. Will be 32 when compiling for x86.
// use popcountll and clzll for 64-bit integers.
}
如果没有
-mpopcnt__builtin_popcountbsr(x) ^ 3131-bsr(x)popcount(~x)32-popcount(x)not32 - popcnt(x) - (31-bsr(x))1 + bsr(x) - popcnt(x)bsrlzcntbsr__builtin_clzstd::countl_zerolzcnt#include <bit>
#include <limits>
int count_significant_zeros_cpp20(unsigned x)
{
return std::numeric_limits<decltype(x)>::digits - std::popcount(x) - std::countl_zero(x);
}
来自 clang 16
-O2 -march=haswellcount_significant_zeros_cpp20(unsigned int):
popcnt eax, edi # missed optimization: false dependency could be avoided with popcnt edi,edi
lzcnt ecx, edi # Haswell also has a false output dependency on lzcnt, although Skylake doesn't.
add ecx, eax
mov eax, 32
sub eax, ecx
ret
如果你想为此使用内联汇编，你可以这样做
int count_significant_zeros_asm(unsigned x)
{
int retval, dummy;
asm(
"lzcnt %[dummy], %[x] \n\t" // false dependency on Intel before Skylake; could xor-zero [dummy] first to break it
"popcnt %[x], %[x] \n\t" // avoid false dependencies on Intel before Ice Lake
"mov %[ret], 32 \n\t"
"sub %[ret], %[dummy] \n\t" // latency of this can overlap with popcnt latency
"sub %[ret], %[x] \n\t"
: [ret]"=r" (retval), [dummy]"=&r"(dummy), // let the compiler pick registers, dummy being an early-clobber output written before we read x for the last time. Although [x] is an RMW inout operand so I don't think it can overlap
[x]"+r"(x) // modify the input reg in place
: // no pure inputs
: // no clobbers
);
return retval;
// dummy is also available with clz(x) in case you want it.
}
如果你想遍历位，用
shr0std::popcount()sbb reg, 0你用
add1顺便说一句，clang 14 及更高版本在
-masm=intelasm("")备选方案：
unsigned int count_nonleading_zero_bits(unsigned int n) {
unsigned int count = 0;
while (n) {
count += 1 - (n & 1);
n >>= 1;
}
return count;
}
x86_64 组装：
count_nonleading_zero_bits:
xor edx, edx
test edi, edi
je .L1
.L3:
mov eax, edi
not eax
and eax, 1
add edx, eax
shr edi
jne .L3
.L1:
mov eax, edx
ret
手动优化（不一定更快）：
count_nonleading_zero_bits:
xor eax, eax ;; set CF=0
.L3:
adc eax, 0
shr edi
cmc
jne .L3
ret
感谢大家不仅提供代码方面的帮助，还提供解释方面的帮助！我真的很感激它。
这是更改后的代码（谢谢 Sep Roland），添加了解释其内部工作的注释。
asm(
"mov eax, %1 ;" // register where binary representation of int x is stored
"mov ebx, 0 ;" // register where the result will be stored
"mov ecx, 32 ;" // loop counter
"loop1:" // check and move to first one in the binary value
"add eax, eax ;" // add value to itself
"jc loop3 ;" // find the first one - jump to loop3 to decrease the counter
"dec ecx ;" // decrement the counter
"jnz loop1 ;" // if counter is not zero, repeat the loop
"jmp loop4 ;" // jump to l4 when eax = 0, when the counter has been zeroed (case where there are no 0s in the register)
"loop2:" // main loop checking the number of zeros
"add eax, eax ;" // add value to itself
"jc loop3 ;" // if carry occurs, jump to loop3
"inc ebx ;" // increment the value holding the number of zeros
"loop3:" // check counter status
"dec ecx ;" // decrement the counter
"jnz loop2 ;" // if counter is not zero, jump to loop2
"loop4:"
"mov %0, ebx ;" // end of operation
: "=r" (y)
: "r"(x)
: "eax", "ebx", "ecx"
);
为了澄清我为什么选择这个。这是因为长度和易于理解对我来说似乎是最好的。我之前说过，我刚刚开始学习这种语言，它与我习惯的高级语言非常不同。
对不起英语，因为它是我的第二语言。