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#2
东海一鱼2010-07-02 17:33
原理：利用补数进行有符号数运算。
.386 ;*********************************************************** code segment use16 assume cs:code,ds:code org 100h ;----------------------------------------------------------- start: jmp short begin msg0 db 'Input minuend : $' ;输入被减数与减数，必须 msg1 db 0dh,0ah,'Input subtrahend : $' ;在-2147483648与 msg2 db 7,'Input error!$' ;2147483647之间 answer db 0dh,0ah,40 dup (' ') buffer db 12,13 dup (0) n1 dd ? n2 dd ? begin: mov dx,offset msg0 call input jc error mov n1,edx mov di,offset answer+2 call hex2dec mov al,'-' stosb mov dx,offset msg1 call input jc error mov n2,edx call hex2dec mov al,'=' stosb mov edx,n1 sub edx,n2 call hex2dec mov al,'$' stosb mov dx,offset answer jmp short ok error: mov dx,offset msg2 ok: mov ah,9 int 21h mov ax,4c00h int 21h ;----------------------------------------------------------- input proc near mov ah,9 int 21h mov dx,offset buffer mov ah,0ah int 21h sub eax,eax mov cl,'+' mov edx,eax mov si,offset buffer+2 ipt0: lodsb cmp al,'-' jne ipt1 mov cl,al jmp ipt0 ipt1: cmp al,'+' je ipt0 cmp al,0dh je ipt2 cmp al,'0' jb ipt4 cmp al,'9' ja ipt4 sub al,'0' shl edx,1 mov ebx,edx shl edx,2 add edx,ebx add edx,eax jmp ipt0 ipt2: cmp cl,'-' jne ipt3 neg edx ipt3: clc jmp short ipt5 ipt4: stc ipt5: ret input endp ;----------------------------------------------------------- hex2dec proc near jmp short h2d0 tmp db 10 dup (' ') h2d0: test edx,80000000h ;检查是否为负 jz h2d1 mov al,'-' neg edx stosb h2d1: push di mov ebx,10 mov eax,edx mov di,offset tmp+9 mov cx,bx h2d2: sub edx,edx div ebx add dl,'0' mov [di],dl or eax,eax jz h2d3 dec di loop h2d2 h2d3: mov si,di pop di sub bx,cx mov cx,bx inc cx rep movsb ret hex2dec endp ;----------------------------------------------------------- code ends ;*********************************************************** end start 以前刚学汇编练手的代码，原作者记不清了。感谢他的代码。  [ 本帖最后由 东海一鱼 于 2010-7-2 17:36 编辑 ] |
R0,R1两个有符号数，同号时两数绝对值相减，异号时两数绝对值相加，并使结果始终与R0符号相同。麻烦详细讲下思路，有代码最好。