汇编语言（assembly language）实现方程的计算

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##汇编语言（assembly language）实现方程的计算

*题目描述：y=4x-5，输入一个十进制一位的x，输出y

其实是有三种做法的。。

2.

data segment m1 db 0dh,0ah,'y=4*x-5;x=','$' m2 db 0dh,0ah,'y=','$' results db ?,?,"$" code segment assume cs:code,ds:data start: mov ax,data mov ds,ax input: lea dx,m1 mov ah,9 int 21h mov ah,1 int 21h cmp al,1bh je exit and al,00001111b ;根据ascii，减去30h mov bl,4 mul bl sub ax,5 mov cx,ax lea dx,m2 mov ah,9 int 21h mov ax,cx cmp ax,8000h ja flag temp: mov bl,10 div bl ;al商,ah余ax商,dx余 or al,00110000b or ah,00110000b cmp al,'0' je s2 jne s1 flag: mov ah,2 mov dl,'-' int 21h mov ax,cx neg ax jmp temp s1: mov results,al mov results+1,ah jmp output2 s2: mov results,ah mov results+1,'$' output2: mov ah,9 lea dx,results int 21h jmp input exit: mov ah,4ch int 21h code ends end start

3

data segment m1 db 0dh,0ah,'y=4*x-5;x=','$' m2 db 0dh,0ah,'y=','$'加粗样式 r0 db 0dh,0ah,'y=-5','$' r1 db 0dh,0ah,'y=-1','$' results db ?,?,"$" code segment assume cs:code,ds:data start: mov ax,data mov ds,ax input: lea dx,m1 mov ah,9 int 21h mov ah,1 int 21h cmp al,1bh je exit cmp al,'1' je output1 cmp al,'0' je output0 and al,00001111b ;根据ascii，减去30h mov bl,4 mul bl sub ax,5 mov bl,10 div bl ;al商,ah余ax商,dx余 or al,00110000b or ah,00110000b cmp al,'0' je s2 s1: mov results,al mov results+1,ah jmp output2 s2: mov results,ah mov results+1,'$' jmp output2 output0: lea dx,r0 mov ah,9 int 21h jmp input output1: lea dx,r1 mov ah,9 int 21h jmp input output2: mov ah,9 lea dx,m2 int 21h mov ah,9 lea dx,results int 21h jmp input exit: mov ah,4ch int 21h code ends end start