Question

我正在尝试逐位减去2个整数，我得到了这个算法

b = 0
difference = 0
for i = 0 to (n-1)

    x = bit i of X
    y = bit i of Y
    bit i of difference = x xor y xor b
    b = ((not x) and y) or ((not x) and b) or (y and b)

end for loop

我已经实施了这一行b = ((not x) and y) or ((not x) and b) or (y and b)

我应该如何在我的代码中实现算法的最后一行

这是我到目前为止所做的：

INCLUDE Irvine32.inc
.data
prompt1 BYTE "Enter the first integer: ",0dh,0ah,0
prompt2 BYTE "Enter the second integer: ",0dh,0ah,0
prompt3 BYTE "The first integer entered is not valid ",0dh,0ah,0
prompt4 BYTE "The second integer entered is not valid ",0dh,0ah,0
X byte 0
Y byte 0
diff byte 0

.code
main PROC

L1:
    mov edx, OFFSET prompt1
    call writeString
    xor edx, edx
    call readInt
    js printError1
    cmp eax, 0ffh
    jg  printError1
    mov X, al
    xor eax, eax

    L2:
    mov edx, OFFSET prompt2
    call writeString
    xor edx, edx
    call readInt
    js printError2
    cmp eax, 0ffh
    jg  printError2
    mov Y, al
    xor eax, eax
    jmp calculation

printError1:
    mov edx, OFFSET prompt3
    call writeString
    xor edx, edx
    jmp L1
printError2:
    mov edx, OFFSET prompt4
    call writeString
    xor edx, edx
    jmp L2

calculation:
mov ebx, 0
mov diff, 0
mov ecx, 7

subtract:
    mov al, X
    and al, 1h
    mov dl, Y
    and dl, 1h
    xor al, dl
    xor al, bl
    mov diff, al




    rol X, 1
    rol Y, 1
    loop subtract
    exit
main ENDP

END main

算法从计算循环标签开始。我需要保存存储在al寄存器中的值，以便实现算法的最后一行，但是由于使用了dl和bl，我应该使用哪个通用寄存器来存储al的值？

提前谢谢

Answer 1

没有你的代码仍然是错的。下面是一段代码，展示了如何在堆栈中存储寄存器。（然而，它远未优化）通常，如果您没有注册，请使用堆栈。如果寄存器在代码中的其他位置使用并需要保留，请使用堆栈存储它们，然后在完成后重置它们。

calculation:
        mov ebx, 0
        mov ecx, 7
subtract:
        ; init
        mov eax, 0
        mov edx, 0

        ; al = bit i of x
        mov al, X
        and al, 1h

        ; dl = bit i of y
        mov dl, Y
        and dl, 1h

        ; save data for later (technique 1 the stack)
        push eax
        push edx

        ; bit i of difference = x xor y xor b
        xor al, dl
        xor al, bl
        or diff, al ; or instead of mov

        ; restore data (technique 1 the stack)
        pop edx
        pop eax

        ; b = ((not x) and y) or ((not x) and b) or (y and b)
        not al
        mov dh, al ; copy not al in dh (technique 2)
        and al, dl ; ((not x) and y)
        and dh, bl ; ((not x) and b)
        and dl, bl ; (y and b)
        or  al, dh ; ((not x) and y) or ((not x) and b)
        or  al, dl ; ((not x) and y) or ((not x) and b) or (y and b)
        mov bl, al

        ror diff, 1
        ror X, 1
        ror Y, 1
        loop subtract
        ror diff, 1

Answer 2

INCLUDE Irvine32.inc
.data
prompt1 BYTE "Enter the first integer: ",0dh,0ah,0
prompt2 BYTE "Enter the second integer: ",0dh,0ah,0
prompt3 BYTE "The first integer entered is not valid ",0dh,0ah,0
prompt4 BYTE "The second integer entered is not valid ",0dh,0ah,0
prompt5 BYTE "The result is: ",0dh,0ah,0
X byte 0
Y byte 0
sum byte 0

.code
main PROC

L1:
    mov edx, OFFSET prompt1
    call writeString
    xor edx, edx
    call readInt
    js printError1
    cmp eax, 0ffh
    jg  printError1
    mov X, al
    xor eax, eax

    L2:
    mov edx, OFFSET prompt2
    call writeString
    xor edx, edx
    call readInt
    js printError2
    cmp eax, 0ffh
    jg  printError2
    mov Y, al
    xor eax, eax
    jmp calculation

printError1:
    mov edx, OFFSET prompt3
    call writeString
    xor edx, edx
    jmp L1
printError2:
    mov edx, OFFSET prompt4
    call writeString
    xor edx, edx
    jmp L2

calculation:
mov ebx, 0
mov bh, 0
mov ecx, 8

subtract:
    mov al, X
    and al, 1h
    mov dl, Y
    and dl, 1h
    mov ah, al
    mov dh, al
    xor al, dl
    xor al, bl
    mov bh, al
    add sum, bh
    not ah
    and ah, dl
    not dh
    and dh, dl
    and dl, bl
    or ah, dh
    or ah, dl
    mov bl, ah

    ror X, 1
    ror Y, 1
    loop subtract

    xor eax, eax
    mov al, sum
    js printError1
    cmp ebx, 0ffh
    jg  printError1
    jmp printResult

    printResult:
        xor edx, edx
        mov edx, OFFSET prompt1
        call writeString
        call writeInt

    exit
main ENDP

END main

好的，我明白了

在汇编中逐位减去两个整数

2 个答案: