org 100h        ; start address (origin) of program.
                ; init vars and registers.
mov cl,128      ; init cl  - 128 is the highest bit value in our byte. This is out comparator.
mov ah,2        ; we use 2 (with int 21h) to output one character (byte) at a time.
start:
;move original to temp register
mov bl, [Val2Cnv]
and bl,cl       ;AND our temp value with comparator (CL)
cmp bl,cl       ;Compare result of AND
je o1           ;Set output as 0
mov dx, 30h
jmp o           ;jump to the actual output
o1:             ;Set output as 1
mov dx, 31h
o:
int 21h         ;output the result we set
shr cl, 1       ;bit shift (effective divide by 2!)
cmp cl, 0       ;check if we're at zero (ie, all done)
ja start
ThatsAllFolks:
;finish up
mov dx,13       ;CR character
int 21h         ;execute the output (ah is 2)
mov dx,10       ;LF character
int 21h         ;execute the output (ah is 2)
mov ax,4c00h    ;exit nicely and tell the OS all is well!
int 21h         ;execute the exit

Val2Cnv db 125  ; the value to convert.
    

BA1300                   MOV DX,13
B213                     MOV DL,13
    

macro putc char {

mov ah,2
mov dl,char
int 21h

}

org 100h

putc 'H'
putc 'i'
putc '!'

int 20h
    

c fasm167.zip

int 21h,2 only uses DL

org 256

xor ecx,ecx

p0:

mov bx,8
mov eax,ecx
inc ecx

p1:
dec bx
mov edx,eax
and dl,1
shr eax,1
or dl,48
mov [b+bx],dl
cmp bx,0
ja p1

mov ah,9
mov dx,b
int 33

cmp ecx,255
jbe p0

int 32
b db 8 dup 48,13,10,36
    

I don't understand the purpose of that first XOR on ecx!

eVAPor8 wrote:

I don't understand the purpose of that first XOR on ecx!

xor ecx, ecx = zero ecx

xor ecx,ecx is assembled to 31 C9
mov ecx,0 is assembled to B9 00 00 00 00
They are both one-cycle once decoded, but it is better to use short instructions so the processor does not use as much time reading code from memory.

    1010    1010
ADD      OR
    1010    1010
   =====   =====
   10100   01010