[Solved] How does the following Assembly language code run?

Question

This code prints a prompt and inputs a number. It then prints that number of lines of stars. On the first line is 1 star, the second line 2 stars, and so on. I have annotated the code to make it more clear.

The code does this with two nested loops. The ecx register is used for both loops: as a counter for the stars on each line, and for the line count. That is why ecx is pushed and popped, so it can have another count in the inner loop.

TITLE MASM Template (main.asm)      ;used for listings etc.

; Description: 
;
; Revision date:f

INCLUDE Irvine32.inc                ;include another code file

.data                               ;data segment

counter dword 1                     ;characters per line
instruct1 BYTE"How many line are required?: ",0
answer BYTE"Answer:",0              ;irrelevant to this code
newline BYTE 0Dh, 0Ah               ;used by crlf
sum BYTE 0                          ;irrelevant to this code

.code                               ;code segment

    main PROC                       ;declare code block

    mov edx,OFFSET instruct1        ;message pointer
    call WriteString                ;display message
    call readint                    ;input an integer
    mov ecx,eax                     ;move input to line loop register
L1:
    push ecx                        ;save line count register
    mov ecx,counter                 ;load character counter
L2:
    mov al,'*'                      ;the char we wish to print
    call writechar                  ;output one char
    loop L2                         ;next character (count in cx)
    pop ecx                         ;restore the line counter
    inc counter                     ;increment characters per line
    call crlf                       ;print the newline defined above
    loop L1                         ;next line (count in cx)

    exit                            ;return to OS
    main ENDP                       ;end code block
    end main                        ;end of code file

If the input is 3, the output will be:

*
**
***

As an aside, I would criticise the author of the code for the following line, for two reasons.

mov ecx,eax     ; move ecx to eax

Reason 1. The comment is back to front; move the returned value eax to ecx for the line counter

Reason 2: Never use a comment to explain what an instruction does, you can RTM for that. Use comments for added value, to make it clear what the purpose is.

Accepted Answer