用汇编语言将32位16进制数转换成10进制数，怎么办？

Question

思路是什么呢？位数这么大……

Answer 1

思路：

把十六进制（比如4位）数除以10000d（十进制数，因为十进制最大是5位--65535）
如果二位十六进制数就除以100d（因为十进制最大是三位--255）
商就是十进制的最高位
再把余数除以1000d
商就是十进制的次高位
再把余数除以100d
...
除以10d
....
除以1d

...以此类推
就得到十进制数

下面是四位十六进制数转为十进制代码：
code segment
main proc far
assume cs:code
start:
call hexibin ;16-2
call crlf ;换行

call binidec ;2-10
call crlf

jmp main ;get next input
ret
main endp

hexibin proc near
mov bx,0 ;clear BX for number
newchar:
mov ah,01h ;keyboard input
int 21h ;call DOS
sub al,30h ;ASCII to binary
jl exit ;jump if<0
cmp al,10d ;is it > 9d
jl add_to ;yes,so it's digit

;not digit(0-9),may be letter(a to f)
sub al,27h ;convert ASCII to binary
cmp al,0ah ;is it <0a hex?
jl exit ;yes,not letter
cmp al,10h ;is it > 0f hex?
jge exit ;yes,not letter
;is hex digit,add to number in BX
add_to:
mov cl,4
shl bx,cl
mov ah,0
add bx,ax
jmp newchar
exit:
ret
hexibin endp

binidec proc near
mov cx,10000d
call dec_div
mov cx,1000d
call dec_div
mov cx,100d
call dec_div
mov cx,10d
call dec_div
mov cx,1d
call dec_div
ret

dec_div proc near
mov ax,bx ;number low half
mov dx,0 ;zero out high half
div cx
mov bx,dx ;remainder into BX
mov dl,al ;quotient into DL
;print the contents of DL on screen
add dl,30h ;convert to ASCII
mov ah,02h
int 21h
ret
dec_div endp

binidec endp

crlf proc near
mov dl,0ah
mov ah,02h
int 21h

mov dl,0dh
mov ah,02h
int 21h
ret
crlf endp

code ends
end start

Answer 2

去网络上找找运算子程序库，

Answer 3

我告诉你思路啊，用C来编，然后反汇编，看看编译器是怎么生成代码的，嘿嘿！本回答被提问者采纳