Question

I use Intel x86 for assembler programming. I’ve got two variables (int), and I want the assembler function to return the biggest. I call the assembler function with a C program, and I’ve got this in the main(), function(1,5).

Here is the assembler code:

            .globl function

            .data
var1:       .long 0
var2:       .long 0

            .text
function:
            movl    4(%esp), %eax
            movl    8(%esp), %ebx

            cmp     %eax, %ebx
            jg      cond1          /*greater, if a < b */
            jl      cond2          /*lower, if a > b */

            movl    var2, %eax

            ret

cond1:
            movl    %eax, var1     /*var1 = a */
            movl    %ebx, var2     /*var2 = b */
            ret


cond2:
            movl    %eax, var2     /*var2 = a*/
            movl    %ebx, var1     /*var1 = b */
            ret

The biggest number will be in %eax (movl var2, %eax). The problem is that
the function always returns the initial number in %eax.
For example, function(1,5) returns “1” instead of “5”.

I don’t understand why the result is wrong.

EDIT : Thanks to your replies, I’ve modified the program thanks to your advice :

  function:
            movl    4(%esp), %eax
            movl    8(%esp), %ebx

            cmp     %eax, %ebx
            jg      cond1          /*greater, if a < b */
            jl      cond2          /*lower, if a > b */
            next:
            movl    var2, %eax
            ret

cond1:
            movl    %eax, var1     /*var1 = a */
            movl    %ebx, var2     /*var2 = b */
            jmp     next

cond2:
            movl    %eax, var2     /*var2 = a*/
            movl    %ebx, var1     /*var1 = b */
            jmp     next

To come-back in function(), I use jmp, is it correct?
It works fine.

Also, how can I improve this code? I use variables because the aim will be to have three numbers and find the median one.

Answer 1

I think you’re confused about what the jg and jl instructions are doing.

From your code, my best guess is that you’re thinking of them as being approximately equivalent to this C code:

if (condition) cond1();

whereas they actually behave like

if (condition) goto cond1;

So, there are three possible control flow paths through your function:

1) If the jg branch is taken:

----caller----.
              |
              v

function:
             movl   4(%esp), %eax
             movl   8(%esp), %ebx

             cmp        %eax, %ebx   
             jg         cond1            /*greater, if a < b */

              |
           branch 
              |
              v

cond1:
             movl       %eax, var1             /*var1 = a */
             movl       %ebx, var2             /*var2 = b */
             ret

              |
   return to  |
<---caller----'

2) If the jg branch is not taken, but the jl branch is taken:

----caller----.
              |
              v

function:
             movl   4(%esp), %eax
             movl   8(%esp), %ebx

             cmp        %eax, %ebx   
             jg         cond1            /*greater, if a < b */
             jl         cond2                   /*lower, if a > b */

              |
           branch 
              |
              v

cond2:
             movl       %eax, var2            /*var2 = a*/
             movl       %ebx, var1            /*var1 = b */
             ret

              |
   return to  |
<---caller----'

3) If neither branch is taken — this is the only path which executes movl var2, %eax:

----caller----.
              |
              v

function:
             movl   4(%esp), %eax
             movl   8(%esp), %ebx

             cmp        %eax, %ebx   
             jg         cond1            /*greater, if a < b */
             jl         cond2                   /*lower, if a > b */

             movl       var2, %eax

             ret

              |
   return to  |
<---caller----'