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I have to take 16 bit as an input as 1234 and tried to display it. But it gives output as 4660. I tried to store digit by digit because when I take input, it will get stored in al as in ASCII form. After that I tried to shift the whole bit in al to left using shift left(SHL) operation which will give me 10 in al if I have inserted 1. After that I inserted second digit and performing shift and rotate operation, I tried to make it in 02 form if the second digit stored is 2. Further, I performed OR operation on 10 and 02 which are stored in register. I repeated the same process for storing lower 8-bit number. But the output is different.

.model small
.stack 100h
.data

.code
    main proc
    mov ax,@data
    mov ds,ax

    ;taking 16 bit number input
    mov ah,01h
    int 21h
    mov bh,al
    mov cl,4
    shl bh,cl

    mov ah,01h
    int 21h
    mov cl,4
    shl al,cl
    mov cl,4
    ror al,cl
    or bh,al

    mov ah,01h
    int 21h
    mov bl,al
    mov cl,4
    shl bl,cl

    mov ah,01h
    int 21h
    mov cl,4
    shl al,cl
    mov cl,4
    ror al,cl
    or  bl,al
    ;taking 16 bit number input


    ;displaying number in dos
    mov     ax,bx
mov     bx,10         
    xor     cx,cx          
  .a: 
    xor     dx,dx          
    div     bx             
    push    dx             
    inc     cx             
    test    ax,ax          
    jnz     .a             
 .b: 
    pop     dx             
    add     dl,"0"         
    mov     ah,02h         
    int     21h            
    loop    .b
exit:
   mov ah,4ch
   int 21h
   main endp
   end main
Ruslan
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Natalie
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    What are you doing with shl/ror? That looks equivalent to `and al, 0Fh` to zero the high 4 bits, equivalent to the more common `sub al, '0'` to convert an ASCII digit to an integer. – Peter Cordes Apr 20 '19 at 19:28
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    Take a look [here](https://stackoverflow.com/questions/55694364/how-to-convert-a-16-bit-number-entered-by-user-to-decimal) This Q/A deals with exactly the same problem **and solves it**. – Sep Roland Apr 21 '19 at 18:59

1 Answers1

3

After converting ASCII digits to integer, you're packing the 4-bit nibbles together into BCD (Bindary Coded Decimal), rather than multiplying by decimal place-values (powers of 10) to make a binary integer.

4660 (decimal) = 0x1234 so you did correctly implement your BCD packing, but that's not what you want to do in the first place.

Left-shift by 4 is equivalent to multiply by 16, not 10.

NASM Assembly convert input to integer? for string->binary integer using the standard
total = total*10 + digit algorithm, starting with the most-significant digit.

That Q&A has 32-bit code, but the algorithm itself is easy enough to implement once you understand it. Or you can search for other Q&As on stack overflow (e.g. using google if the built-in search doesn't help) with 16-bit code.

Peter Cordes
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