Trying to print asteric triangle in ASSEMBLY Language

Question

.model small
.stack 100h
.data
space db 10,"$"
asteric db "*$"
.code
start:
mov cl,5
mov ax,@data
mov ds,ax
mov bl,0
mov al,0
l1:
mov ah,09
lea dx,space
int 21h
inc bl
cmp bl,cl
je exit
l2:
lea dx,asteric
int 21h
cmp al,bl
jl l2
inc al
jnl l1
exit:
mov ah,4ch
int 21h
end start

i don't know what i'm doing wrong here but it is giving me output like this OUTPUT:

*
*
*
*

i want output like this: OUTPUT:

*
**
***
****

instead of what i'm getting please help

"Assembly" is not a programming language. Which ISA and assembler are you using? — tadman, Mar 29 '18 at 16:57
Presumably for x86? That's the important part. There's a million CPUs out there and they all have their own assemblers. — tadman, Mar 29 '18 at 16:58
Learn to use the nice turbo debugger. Also, comment your code, especially if you want others to help. PS: `10` is not space, it's line feed. — Jester, Mar 29 '18 at 17:05
Also I am surprised you get that output, given that you have an endless loop at `l2`. — Jester, Mar 29 '18 at 17:14
@Jester "space" is just the nomenclature: the output contains newlines but no spaces. To the OP: printing a newline *first* is bad practice, similar to this in C with `printf("\n****");` for example, which is better as `printf("****\n");` — Weather Vane, Mar 29 '18 at 17:31

Tomer Shinar · Answer 1 · 2018-04-11T10:35:58.147

-3

you need to put zero in al before you go into l2 loop(before the lable l2) because if not it will count all of the times you printed before. you also need to inc al before you do jl to l2 because you need to count how many times you printed in this line and in the way you did it it count lines.

edited Apr 11 '18 at 10:35

answered Apr 10 '18 at 15:09

Tomer Shinar

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Trying to print asteric triangle in ASSEMBLY Language

1 Answers1