x86:计数以32位数字从1转换为0 [英] x86: Count transitions from 1 to 0 in 32 bit number

问题描述

我喜欢以32位数字计算从1到0的过渡数.我发现使用shr以相反的顺序从0过渡到1的次数，但没有得到所需的结果.我该怎么办?

I like to count the number of 1 to 0 transition in a 32 bit number. I found that the number of transitions from 0 to 1 in reverse order using shr, but I don't get the required result. What should I do?

extern printf                   
SECTION .data                   
    msg:      db "The number of transitions are : %d",10,0  
    inta1:    dd 1234567   ; integer 1234567        
    num:      dd 4660  
SECTION .text                    

global main               
main:     
    mov eax, [num]    
    mov ecx,32    
    mov ebx,0    
.loop:  dec ecx  
    cmp ecx,0  
    jl .exit   
    shr eax,1  
    jc .loop  
    dec ecx  
    shr eax, 1  
    jnc .loop  
    inc ebx  
    jmp .loop  
.exit:  
    push ebx  
    push    dword msg          
    call    printf             
    add     esp, 8  

输出:

过渡数量为:2

The number of transitions are : 2

而对于4660(00000000000000000001001000110100)，从1到0的转换数是4.

whereas for 4660 (00000000000000000001001000110100) the number of 1 to 0 transitions are 4.

推荐答案

基本问题是您的代码将查找0后跟1，如果不匹配，则会重新开始.因此，如果看到00，它将重新开始寻找另一个0；如果下一位是1，它将重新开始.因此，当过渡出现偶数0 s

The basic problem is that your code looks for a 0 followed by a 1 and if it doesn't match, it starts over again. So if it sees 00 it starts over looking for another 0 and if the next bit is 1 it starts over. So you miss transitions when they are preceeded by an even number of 0s

请注意，您还说要从1到0的过渡，但是要从0到1的过渡，但是从右到左(从lsb到msb)，可能是正确的，具体取决于您想要的位顺序.

Note also you say you want 1-to-0 transitions, but you're looking for 0-to-1 transitions, BUT you are looking right to left (lsb to msb) so that may be correct, depending on which bit order you want.

显而易见的解决方案是将jnc .loop更改为不重新开始，而只返回到紧接的前一个dec ecx，尽管那里也需要cmp ecx,0; jl .exit.

The obvious solution is to change the jnc .loop to not start over, but only go back to the immediately preceeding dec ecx, though you also need a cmp ecx,0; jl .exit there.

请注意，当结果(在eax中)全部为0时，您可以使用shr指令设置的Z标志来摆脱对ecx的需要，并计数位数.位.

Note also you could get rid of the need of ecx and counting the number of bits by using the Z flag which is set by the shr instruction when the result (in eax) is all 0 bits.

在一起，您会得到类似的东西

Taken together, that gives you something like:

.loop1: shr eax, 1
        jc .loop1
.loop2: jz .exit
        shr eax, 1
        jnc .loop2
        inc ebx
        jmp .loop1

这篇关于x86:计数以32位数字从1转换为0的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！