Question

我的问题是它没有比较第一个数字是否大于第二个数字..它只执行我打印字符串的部分

    mov ah,9
    lea dx,str1
    int 21h           ;Write string at DS:DX to standard output

    mov ah,1
    int 21h           ;Read character from standard input into AL
    sub al,30h        ;al = character - '0'
    mov num1,al       ;num = character - '0'

    mov ah,9
    lea dx,str2
    int 21h           ;Write string at DS:DX to standard output

    mov ah,1
    int 21h           ;Read character from standard input into AL

    mov al,2

    cmp num1,al       ;Is num1 greater than 2?
    jg sum            ; yes, goto sum
                      ; no  
    mov ah,9
    lea dx,str3
    int 21h           ;Write string at DS:DX to standard output
    jmp exit


sum:
    add num1,al       ;num1 = num1 + AL = num1 + 2

    mov ah,9
    lea dx,new
    int 21h           ;Write string at DS:DX to standard output

    mov ah,9
    lea dx,num1
    int 21h           ;Write string at DS:DX to standard output
    jmp exit

exit:

Answer 1

你写过这个

cmp num1,al       ;Is num1 greater than 2?
jg sum            ; yes, goto sum
                  ; no

你正在接受cmp语句的错误含义

正确的含义和陈述（对于这个场景）是

cmp num1,al       ;Is 2 lesser than num1?
    jl sum            ; yes, goto sum
                      ; no

我修改了你的代码，现在它就像这样

    org  100h
    jmp start
    str1 dw 'Provide Input',10,13,'$' 
    str2 dw 'Provide another input',10,13,'$' 
    str3 dw 'First is greater',10,13,'$'
    new dw 'Second is greater',10,13,'$'   
    num1 db ?
    start:
    mov ah,9

    lea dx,str1
    int 21h           ;Write string at DS:DX to standard output

    mov ah,1
    int 21h           ;Read character from standard input into AL
    sub al,30h        ;al = character - '0'
    mov num1,al       ;num = character - '0'

    mov ah,9
    lea dx,str2
    int 21h           ;Write string at DS:DX to standard output

    mov ah,1
    int 21h           ;Read character from standard input into AL
    sub al,30h        ;al = character - '0'

    cmp num1,al       ;Is al lesser than num1?
    jl sum            ; yes, goto sum
                      ; no  
    mov ah,9
    lea dx,str3
    int 21h           ;Write string at DS:DX to standard output
    jmp exit


sum:
    add num1,al       ;num1 = num1 + AL = num1 + al

    mov ah,9
    lea dx,new                
    int 21h           ;Write string at DS:DX to standard output

exit:
    MOV AH,4CH 
    INT 21H

现在尝试以这种方式首先输入2比5，请参阅输出。现在再次运行它并输入5比2的输入。当你自己运行它时，你会了解更多。

我制作了一个程序，只有当第一个数字大于第二个数字并且不进行比较时才会添加两个数字

1 个答案: