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

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.