组装，打印ASCII码 [英] Assembly, printing ascii number

问题描述

我的汇编代码有问题.我想打印存储在寄存器 cx 中的数字，但是当我尝试打印它时，它打印的是 ascii 字符而不是 ascii 数字，所以我决定编写一个程序将 ascii 字符转换为 ascii 值.问题是，当我尝试调用该过程时，程序会冻结，我必须重新启动 dosbox.有谁知道这段代码有什么问题?谢谢.

I have a problem with my assembly code. I want to print number stored in register cx, but when i tried to print it, it printed ascii character instead of ascii number, so I decided to write a procedure to convert ascii char to ascii value. Problem is, that when I try to call that procedure, the program freezes and I have to restart dosbox. Does anyone know whats wrong with this code? Thanks.

P4      PROC                
            MOV AX,CX           ;CX = VALUE THAT I WANT TO CONVERT
            MOV BX,10           
    ASC2:
            DIV BX              ;DIV AX/10
            ADD DX,48           ;ADD 48 TO REMAINDER TO GET ASCII CHARACTER OF NUMBER 
            PUSH AX             ;SAVE AX
            MOV AH,2            ;PRINT REMAINDER STORED IN DX
            INT 21H             ;INTERRUP
            POP AX              ;POP AX BACK
            CMP AX,0            
            JZ EXTT             ;IF AX=0, END OF THE PROCEDURE
            JMP ASC2            ;ELSE REPEAT
    EXTT:
            RET
    P4      ENDP

推荐答案

这样的方法更适合打印十进制值(新代码为小写):

Something like this would work better for printing a decimal value (the new code is in lowercase):

        mov byte [buffer+9],'$'
        lea si,[buffer+9]

        MOV AX,CX           ;CX = VALUE THAT I WANT TO CONVERT
        MOV BX,10         
ASC2:
        mov dx,0            ; clear dx prior to dividing dx:ax by bx
        DIV BX              ;DIV AX/10
        ADD DX,48           ;ADD 48 TO REMAINDER TO GET ASCII CHARACTER OF NUMBER 
        dec si              ; store characters in reverse order
        mov [si],dl
        CMP AX,0            
        JZ EXTT             ;IF AX=0, END OF THE PROCEDURE
        JMP ASC2            ;ELSE REPEAT
EXTT:
        mov ah,9            ; print string
        mov dx,si
        int 21h
        RET

buffer: resb 10

不是直接打印每个字符，而是以相反的顺序将字符添加到缓冲区.对于值 123，它将在缓冲区 [8] 处添加 '3'，在缓冲区 [7] 处添加 '2'，在缓冲区 [6] 处添加 '1' - 因此，如果您随后打印从缓冲区 + 6 开始的字符串，您将得到123".
我使用的是 NASM 语法，但希望它应该足够清楚.

Instead of printing each character directly it adds the characters to a buffer in reverse order. For the value 123 it would add '3' at buffer[8], '2' at buffer[7] and '1' at buffer[6] - so if you then print the string starting at buffer+6 you get "123".
I'm using NASM syntax but hopefully it should be clear enough.

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