CODEWITHSUNDEEP
assemblyx86nasm
user3561699
user3561699

Reputation: 3

Order of execution in this assembly code? (NASM)

I tried searching for an answer to this and the examples with any similarity at all were either too simple or too complex. I'm using Paul Carter's book for an assembly class, so some of the macros are from him. Namely, the print_string, print_int, and print_nl functions.

Given this snippet of assembly:

segment .data

        output db "Welcome!",0
        string1 db "greater than",0
        string2 db "less than",0
        string3 db "equal to",0
        var_a dd 0Ah

segment .bss

segment .text
        global  _asm_main

_asm_main:

        enter   0,0               ; setup routine
        pusha
    ;***************CODE STARTS HERE***************************

        mov eax, output
        call print_string
        call print_nl

        cmp dword[var_a], 0Ah
        jle label1
        mov eax, string1
        call print_string
        call print_nl
        jmp label3

    label1:
        cmp dword[var_a], 9h
        jg label2
        mov eax, string2
        call print_string
        call print_nl
        jmp label3
    label2:
        mov eax, string3
        call print_string
        call print_nl
    label3:
        cmp dword [var_a], 0
        jle label4
        sar [var_a], 1
        mov eax, dword [var_a]
        call print_int
        call print_nl
        jmp label3
    label4:

    ;***************CODE ENDS HERE*****************************
        popa
        mov     eax, 0            ; return back to C
        leave
        ret

The output is:

Welcome!

equal to
5
2
1
0

My question:

I see how it eventually gets to label2 via the comparisons, prints the "equal to" message and a new line. After that though, I see no additional comparisons in label2. At that point, how does label3 even get invoked? How does the program "leave" label2? I understand everything in the assembly except in between the end of label2 (which prints the "equal to") and the start of label3 (where it loops a bit shift and prints the numbers). Am I missing something?

After label2 is executed, does control of the program go back to where label2 was first called in label1? Or does label3 execute because it's next sequentially?

Upvotes: 0

Views: 1526

Answers (1)

Nayuki
Nayuki

Reputation: 18552

Imagine that your code had no labels at all. The machine would execute instructions one by one, going down the list. Adding labels does not change this behavior at all; the machine will still go to the next instruction unless there's a jump.

A label gives a name to a memory address. They are useful when jumping. If we didn't have labels, then we might need to do something like this:

add foo, bar
sub foo, bar
mul foo, bar
jmp -2         ; Fake code for "go up 2 instructions"

With labels, we can give names to the memory address of an instruction:

  add foo, bar
thingamajig:       ; This label equals the address of the 'sub' instruction
  sub foo, bar
  mul foo, bar
  jmp thingamajig

Upvotes: 2

Related Questions