How to convert a 16 bit number entered by user to decimal

问题

I got the user to enter a 16 bit number. I want to display the number entered by the user.

This is what I've come up with so far.

As you can see, I have subtracted 30H from the input to convert it to decimal. Where should I add 30H back to get it back to its original ASCII value?

        MOV AH,1H         ;user input for first number part 1 
        INT 21H 
        SUB AL,30H  
        MOV NUM1,AL

        MOV AH,1H         ;1st number part 2
        INT 21H           
        SUB AL,30H 
        MOV NUM2,AL  

        MOV AH,1H         ;1st number part 3
        INT 21H
        SUB AL,30H
        MOV NUM3,AL                         

        MOV AH,1H         ;1st number part 4
        INT 21H
        SUB AL,30H 
        MOV NUM4,AL

        XOR AH,AH                  
        MOV AL,NUM1       
        MOV DX,1000D
        MUL DX            ;1*1000
        ADD AH,30H
        ADD AL,30H
        MOV BX,AX        

        XOR AH,AH                  
        MOV AL,NUM2
        MOV DX,100D       
        MUL DX            ;2*100
        ADD BX,AX

        XOR AH,AH
        MOV AL,NUM3
        MOV DX,0010D
        MUL DX            ;3*10
        ADD AH,30H
        ADD AL,30H
        ADD BX,AX  

        XOR CH,CH
        MOV CL,NUM4
        ADD CH,30H
        ADD CL,30H
        ADD BX,CX       ;BX now has the 16 bit number


        MOV FNUM1,BX      ;final 1st 16 bit number


        PRINTN

        LEA DX,MSG8       ;msg for output
        MOV AH,9H
        INT 21H

        PRINT FNUM1

For eg: when I enter the number as 1234, I get output as F. Can you guys let me know what I'm doing wrong and help me out? Any help would be appreciated at this point.

回答1:

First get the input right.

Once you got the 4 digits from input you need to combine them adhering to the formula
d1 * 1000 + d2 * 100 + d3 *10 + d4 Nowhere in this calculation are you required to add in 48!

The multiplication *1000 needs a word-sized mul, but multiplying *100 and *10 can do with a byte-sized mul.

mov     al, NUM1
cbw                   ; -> AH=0
mov     dx, 1000
mul     dx            ; Product 1*1000 in DX:AX but with DX=0
mov     bx, ax        

mov     al, 100
mul     NUM2          ; Product 2*100 in AX
add     bx, ax

mov     al, 10
mul     NUM3          ; Product 3*10 in AX
add     bx, ax  

add     bl, NUM4
adc     bh, 0         ; BX now has the 16 bit number

A nicer version of the above uses a loop. This loop can only work if you define your NUMx variables as consecutive bytes in memory!

NUM1 db 0
NUM2 db 0
NUM3 db 0
NUM4 db 0

...

    mov     cx, 4    ; Number of digits
    lea     si, NUM1 ; Address of 1st digit (most significant digit)
    xor     bx, bx   ; The 16-bit result
More:
    imul    bx, 10   ; BX = BX * 10
    lodsb            ; NUM1 then NUM2 then NUM3 then NUM4
    cbw
    add     bx, ax
    dec     cx
    jnz     More

Then output to the screen.

I want to display the number entered by the user.

Move the 16-bit number stored in BX (from the previous steps) to the AX register and then read my explanations on how to convert a 16-bit number in AX into text so it can be printed to the screen at Displaying numbers with DOS.

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来源：https://stackoverflow.com/questions/55694364/how-to-convert-a-16-bit-number-entered-by-user-to-decimal