Converting decimal to binary in assembler

Question

I need help with my first program in assembler. I have to convert values entered by user from decimal to binary. I have no idea how can I show values as a decimal, and what shou

Answer 1

Not entirely clear what you're trying to do. I guess "decimal to binary", but the prompt says "Enter binary value". I would take that to mean a string of "1"s and "0"s. I wouldn't ask 'em for a "decimal" value either - you'll get like "1.23", which you aren't equipped to handle. Just ask 'em for a number. Maybe "a number less than 65536", which is probably(?) what you want.

Warning! Untested code ahead!

    Number proc
    mov cx,5 ; loop counter?
    xor bx,bx ; "result so far"?


    read:
    mov ah,0
    int 16h

; wanna give 'em the option to enter
; less than the full five digits?
    cmp al, 13 ; carriage return
    jz finis

    cmp al,'0'
    jb read
    cmp al, '9'
    ja read
    mov ah,0eh
    int 10h
; Assuming al still holds your character...
    sub al, '0' ; convert character to number
    mov ah, 0 ; make sure upper byte is clear
    imul bx, bx, 10 ; multiply "result so far" by 10
    ; jc overflow ;  ignore for now
    add bx, ax ; add in the new digit
    ; jc overflow ; ignore for now

    loop read
finis:
; now our number is in bx
; it is conventional to return values in ax
    mov ax, bx

overflow: ; I'm just going to ignore it
    ; spank the user?
    ; go right to exit?
    ret ; maybe endp generates this. two shouldn't hurt
    Number endp

Now I guess you want to print the binary ("1"s and "0"s) representation of that number...

    Printbin proc
; what does "proc" do? Do you know?
    mov bx, ax
    mov cx, 16 ; 16 bits to do, right?
    mov ah, 2
 top:
    mov dl, '0'
    shl bx, 1 ; shift leftmost bit to carry flag
    adc dl, 0 ; bump the "0" up to "1", if set
    int 21h
    loop top
    ret

    endp ; ?

Been a while since I've done DOS, so there may be serious errors in that, but it may give you some ideas.

Answer 2

I think it will be ok for you.

; Read an integer from the screen and display the int in binary format
; and continue until number is negative.
again:            ; For loop
    call read_int ; take the integer from screen
    cmp eax,0     ; look if number is not negative
        JL end:       ; if less than zero program ends.
    mov ecx,32    ; for loop we set ecx to 32 ; ATTENTION we not specified type. So compiler will get error.

    mov ebx,eax   ; we will lost our number in eax, so I take it to ebx
START:
    xor eax,eax   ; eax = 0
    SHL ebx,1     ; shift the top bit out of EBX into CF
    ADC eax,0     ; EAX  = EAX + CF + 0 ADD CARRY FLAG, so eax is zero we add zero. The new eax will exact value of Carry Flag which is out bit.
    call print_int ; Then we print the CF which we took the eax.
LOOP start:   ; Loop looks ecx if not 0 it goes start.  
call print_nl ; For next number we print a new line
JMP again:    ; For take new number

    END:      ; End of the program.

setc al would also work instead of adc eax,0, and be more efficient on some CPUs.