问题描述
我正尝试逐位减去2个整数，并为此算法提供了
I'm trying to subtract 2 integers bit-by-bit and was given this algorithm
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
我已经实现了这一行 b =((not x)and y)或((not x)and b)或(y and b).我应该如何在代码中实现算法的最后一行
I have implemented up to this line b = ((not x) and y) or ((not x) and b) or (y and b). How should I implement that last line of the algorithm in my code
这是我到目前为止所拥有的:
This is what I have so far:
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.为了实现算法的最后一行，我需要保存存储在 al 寄存器中的值，但是由于 dl 和 bl 在我应该使用哪个通用寄存器来存储 al 的值?
The algorithm start from the calculation loop label. I needed to save the value stored in al register, in order to implement the last line of the algorithm, but since dl and bl is in used which general purpose register should I use to store value of al?
推荐答案
您的代码仍然没有错.以下是一段代码，显示了如何在堆栈中存储寄存器.(但是，距离优化还很远)通常，如果您的寄存器不足，请使用堆栈.如果寄存器在代码的其他位置使用过并且需要保留，请使用堆栈存储它们，然后在完成后将其重置.
No your code is still wrong. Below is a piece of code that shows how to store registers in the stack. (It is however far from optimized)In general if you’re out of registers, use the stack.If registers are used in other places in your code and need to persist, use the stack to store them then reset them when you’re done.
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
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