Question

我有一个任务，使用NASM程序集找到数字中最不重要和最重要位的位置。但是，我遇到了两个问题，即使得到我要编译的东西：

将数字除以2不起作用（div 2输出invalid operand错误）
我的注册用完了！

这就是我现在所拥有的：

%include "along32.inc"

section .data
msg1    db      'Enter a hexadecimal number: ', 0
msg2    db      'LSB set: ' , 0
msg3    db      'MSB set: ', 0
msg4    db      'Total bits set: ', 0


section .text

global main

main:
    mov     edx,    msg1
    call    WriteString
    call    ReadHex     ; eax contains hex number
    mov     edx,    eax ; edx contains hex number

    mov     si,    0   ; si will contain lsb posiiton
    mov     r9w,    0   ; r9w will contain msb position
    mov     bh,     0   ; bh will contain number of bits set
    mov     bl,     1   ; bl will store if the msb is set
    mov     cx,     32  ; loop through every power of two
    mov     eax,    2147483647 ; eax contains 2^31 - 1

bitchecker:
    mov     ebp,    edx ; ebp contains number
    and     ebp,    eax ; ebp contains eax & ebp
    jz      loopend
    inc     bh
    mov     si,    cx ; put value of cx into si
    cmp     bl,     0  ; if msb is already set, jump to loop end
    je      loopend

    mov     r9w,    cx ; put value of cx into r9w
    dec     bl

loopend:
    xor     edx,    edx
    div     2
    loop    bitchecker

continue:
    mov     edx,    msg2
    call    WriteString

    mov     eax,    si ; write lsb
    call    WriteInt

    call    Crlf

    mov     edx,    msg3
    call    WriteString

    mov     eax,    r9w ; write msb
    call    WriteInt

    call    Crlf

    mov     edx,    msg4
    call    WriteString 

    xor     eax,    eax
    mov     ax,    bh ; write total bits set
    call    WriteInt

    jmp end

end:

    call    Crlf

    mov     eax,    1
    int     0x80

我甚至不知道逻辑是否首先起作用，但现在这不是我的问题。这只是很难与所有这些寄存器一起工作，这让人感到困惑。我该如何避免这些问题？

编辑：

最终解决方案：

%include "along32.inc"

section .data
msg1    db      'Enter a hexadecimal number: ', 0
msg2    db      'LSB set: ' , 0
msg3    db      'MSB set: ', 0
msg4    db      'Total bits set: ', 0

section .text

global main

main:
    mov     edx,    msg1
    call    WriteString
    call    ReadHex     ; eax contains hex number

    mov     ecx,    0  ; ecx contains loop variable
    mov     ebx,    32  ; ebx contains lsb
    mov     ebp,    0   ; ebp contains msb
    mov     esi,    0   ; esi contains how many bits set
    jmp     algo

algo:
    shr     eax,    1
    jnc     not_set
    mov     ebp,    ecx
    inc     esi
    cmp     ecx,    ebx
    jg      not_set
    mov     ebx,    ecx

    ; decrement and check loop condition
not_set:
    inc     ecx
    cmp     ecx,    32
    je      end

    jmp     algo


end:
    mov     edx,    msg2
    call    WriteString

    mov     eax,    ebx
    call    WriteInt

    call    Crlf

    mov     edx,    msg3
    call    WriteString

    mov     eax,    ebp
    call    WriteInt

    call Crlf

    mov     edx,    msg4
    call    WriteString

    mov     eax,    esi
    call    WriteInt

    call    Crlf

    mov     eax,    1
    int     0x80

Answer 1

您可以使用div 2。

，而不是无效的shr eax, 1

一个简单的算法是：

lsb = 32;
for(i = 0; i < 32; i++)
{
    if (x & 1)
    {
        if (i < lsb) lsb = i;
        msb = i;
    }
    x >>= 1;
}

这只需要4个寄存器（x，i，lsb和msb）。

汇编：获取LSB和MSB的位置

1 个答案: