Well, recently I started learning Assembly 8086 out of curiosity mostly.
Input in assembly allows you to type only one character, so I tried to make a program in Assembly 8086 that allows you to enter multi-digit integer input, ending the input with "space"(' '), then adding the number and printing the value.
I saw that push and pop can be used to pass arguments to a procedure but I tried to use them to make my procedure return something and store it to a variable, I couldn't imagine a way to do this with ret based on my knowledge on Assembly 8086 so...anyway, I made a procedure, but for some reason the ret at the end of the procedure doesn't seem to work and the procedure runs infinite times.
The code so far:
.model small
org 100h
.data
fv db 0 ;variables to store the numbers
sv db 0 ;
.code
jmp start ;a pattern of mine in some way to avoid a procedure be called twice and what everyone shows
;just doesn't work, after the main ends, all the procs run again,
;it worked perfectly any other time I used procedures to my program
f1 proc ;the procedure
mov cl, 0 ;clear cl bcs later the first time is used I have not stored some thing in there and
;always for some reason, to all my programs "cx" has a value stored, maybe from the
;emulator I use
mov ah, 1h ;single character input
int 21h ;
while:
cmp al, ' ' ;if input is equal to "space"("space" must be the last input, as far as I have
;gone with the program)
jne true ; if input != ' '
je false ; if input == ' '
true: ; in case input != ' '
mov bl, al ;store the input
mov al, cl ;digits taken from input previously
sub bl, 30h;bcs if input == 8, whatactually is stored is the ASCII code of it in this
;case : 38h or 56
mov dl, 10 ;What I thought : if user writes as input 12, what actually types 1 * 10 + 2
;so I am storing 10 to dl to multiply the previously entered numbers in the
;example above : 1
mul dl ;multiplication
add al, bl ;add new input to (old inputs * 10)
mov cl, al ;store old inputs
mov ah, 1h ;input
int 21h ;
jmp while ;check again if input == ' ' or not
false: ;in case input == ' '
mov ch, 0 ; in chase ch had some thing else in it from something else than the
; input(0 <= input <= 127~128(127 + 128 = 255))
push cx ; store cx(final result from the inputs) in to the stack to store it to a
; variable
ret ; end procedure
f1 endp ;
start: ; with "jmp start" at the start of the ".code" makes the program start from "main"
main proc
call f1 ;call procedure
pop bx ;store result in to bx bcs `push` and `pop` as far as I know need at least 16-bit
;and varables are 8-bit, I know I can make them 16-bit but anyway
mov fv, bl ;store result to variable
endp
end main
Well, I found it, before I push anything else to the stack(I searched how push, pop, call and ret work)I pop-ed in to bx and then after I push-ed to the stack what I wanted to push, before ret I push-ed what was to bx(the addres ret was supposed to jump) and then I ret.
Code :
.model small
org 256
.data
fv db 0
sv db 0
.code
jmp start
f1 proc
mov cl, 0
mov ah, 1h
int 21h
while:
cmp al, ' '
jne true
je false
true:
mov bl, al
mov al, cl
sub bl, 30h
mov dl, 10
mul dl
add al, bl
mov cl, al
mov ah, 1h
int 21h
jmp while
false:
pop bx
mov ch, 0
push cx
push bx
ret
f1 endp
start:
main proc
call f1
pop bx
mov fv, bl
endp
end main