int 21h ah 02h doesn't work for some reason

Question

My task is to read a number from a terminal (aka atoi) and then write it back to terminal (aka itoa).

To read a string I use int 21h, ah 0ah. It seems nice when I check it in the debugger. And then my atoi and itoa also look nice except that in itoa I use int 21h, ah 02h to display one character, but for some reason it is not displayed. So I went back to the beginning and noticed that writing int 21h, ah 02h (print char) right after int 21h, ah 0ah (read string) gives no result.

; STACK SEGMENT
STSEG SEGMENT PARA STACK "STACK"
    DB 64 DUP (0)
STSEG ENDS

; DATA SEGMENT
DSEG SEGMENT PARA PUBLIC "DATA"
    startStr db "Enter a number : $"
    inputError db "Number has incorrect chars.", 10, "$"
    buffError db "Number is too big", 10, "$"
    bufferSize DB 16        ; 15 chars + RETURN
    inputLength DB 0        ; number of read chars
    buffer DB 16 DUP('?')   ; actual buffer
DSEG ENDS

; CODE SEGMENT

CSEG SEGMENT PARA PUBLIC "CODE"

    MAIN PROC FAR
        ASSUME cs: CSEG, ds: DSEG, ss:STSEG
        mov ax, DSEG
        mov ds, ax
        ;lea dx, startStr
        ;call WRITING
        ;call READING

        xor ax, ax
        xor dx, dx
        mov ah, 2
        mov dl, 56
        int 21h

;test display char before reading string - works fine


        xor dx, dx
        xor ax, ax
        lea dx, bufferSize
        mov ah, 10
        int 21h

; reading string - works fine

        xor ax, ax
        xor dx, dx
        mov ah, 2
        mov dl, 56
        int 21h

; trying to display char again - nothing

        ; call ATOI
        ; add ax, 15
        ; call ITOA
        ret
    MAIN ENDP
CSEG ENDS
END MAIN

Screenshoot
'8' is the char I've displayed in the beginning. '123' is a string I've entered and then hit enter.

It does work but it's overwriting the one already printed. Use a different character and you will see. — Jester, Sep 30 '19 at 21:52
E.g. change your **second** `mov dl, 56` to `mov dl, 57` then you should see `9` being printed. This is a side effect of the carriage return used to terminate the input. — Jester, Sep 30 '19 at 21:57
I've tried change it to 57 - nothing changes. I've tried change it to 10 (it must be \n) - no result. Can there be any other thing? — Nikolay Zhurba, Sep 30 '19 at 22:01
@Jester yeap. As I said, i've tried different values but nothing changes on my machine — Nikolay Zhurba, Sep 30 '19 at 22:26

score 2 · Accepted Answer · answered Oct 01 '19 at 15:00

Once the DOS buffered input function 0Ah receives the enter key, depending on how your DOS emulator works, will the cursor move to the start of the current line OR to the next line.

Since outputting that second "8" is the only thing left before your program exits, the DOS prompt could be overwriting that second "8". See the screenshot.

Try deferring terminating the program. Just wait for a key.
Also print something different from your first character.

    mov dl, "*"
    mov ah, 02h  ; DOS.PrintChar
    int 21h
    mov ah, 00h  ; BIOS.GetKey
    int 16h

    ret
    MAIN ENDP
CSEG ENDS
END MAIN

int 21h ah 02h doesn't work for some reason

1 Answers1