8086 assembly program function 0ah int 21h

Question

Program should accept a character and an odd number (max 21).

My problem is that it does not seem to accept inputs properly as it always jump to the show error part. Whats wrong with my code?

  .model small
    .stack
    .386
    .data
        msg db "Input a character and an odd number(max 21): $"
        msg1 db 10,13,"ERROR: Input an odd number number after the character", 10, 13, '$'
        msg2 db 10,13, "ERROR: Number is not an odd number", 10, 13, '$'
        inp db ?,?,?
    .code

    main proc far
    again:  mov ax,@data
        mov ds, ax

        mov ah, 9
        lea dx, msg
        int 21h

        lea dx, inp
        mov ah, 0Ah
        int 21h

    ;*********number check******
        mov bl, inp+1
        or bl, 30h ; add 30h 
        cmp bl,30h ; '0'
        jnge notnum   ;  <-------PROGRAM ALWAYS GO HERE FOR ANY INPUT I MAKE
        cmp bl,39h ; '9'
        jnle notnum

        mov bl, inp+2
        cmp bl, 00h
        je numIsOneDigit
        or bl, 30h
        cmp bl,30h ; '0'
        jnge notnum
        cmp bl,39h ; '9'
        jnle notnum
    ;***************************

    ;****if odd number check****
    ;numIsTwoDigit
        mov cl, inp+2
        shr cl, 1
        jnc notOdd
        jmp drawDiamond


    numIsOneDigit:  
        mov ch, inp+1
        shr ch, 1
        jnc notOdd
        jmp drawDiamond
    ;***************************

    ;********ERRORS*************
    notnum:
        mov ah, 9
        lea dx, msg1
        int 21h

        jmp again

    notOdd:
        mov ah,9
        lea dx, msg2
        int 21h

        jmp again
    ;************************
    drawDiamond:
        ;some code here
    exit:
        mov ah, 4ch
        int 21h

    main endp
    end main

TESTING THE PROGRAM:

Answer 1

inp db ?,?,?

Since you want to use the DOS input function 0Ah you need to provide the correct input structure. You defined this structure to only have 3 uninitalized bytes, but DOS expects the first byte to hold the buffer length and the second byte to be reserved so it can return you the number of characters that were actually inputted.

Knowing that you want to input a maximum of 3 characters, the correct structure is:

inp db 4,0,4 dup (0)

It's 4 (not 3) because DOS will also append a carriage return code (13) at the end of the input string. This extra byte will not be included in the count you recieve in the second byte!

---

Having corrected this, you'll need to change the rest of the program also!

;*********number check******
mov bl, inp+1
or bl, 30h ; add 30h 
cmp bl,30h ; '0'

The first digit (if any) will be at inp+3. The second digit (if any) will be at inp+4. These digits are already characters, so I don't understand why you wrote or bl, 30h several times.

mov bl, inp+2
cmp bl, 00h
je numIsOneDigit

If this compare wants to check if a second digit is available then you need to compare with 13 (not with 0).