我做了一個程序，只會在第一個數字大於第二個數字時纔會添加兩個數字，而不是比較

Question

我的問題是它不會比較第一個數字是否大於第二個數字。它只執行在這裏我打印

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

您寫這

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

您正在CMP聲明的錯意字符串的一部分

正確的意義和聲明（FO R此之情況）是

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

我已經修改您的代碼，現在會是這樣

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 

現在嘗試用這種方法首先給輸入2比5，看輸出。現在再次運行並輸入5以上的值2.當你自己運行時，你會明白更多。