Function Compiles Which Arguments First?

A teacher posed a question in class earlier asking if a function that three arguments, which is read first, second, and third (assuming that there are three of them). I've read that it depends on the compiler, but can't seem to get any farther than that. Does anyone have any more input?

say again...?

Ok let try to explain myself once agian. . .

function(int x, int y, int z)
{

. . .

}

Which argument is recognized first, second, and third? Apparently it depends on the compiler, but my teacher wishes for us to write a program/experiment proving this one way or another. Any suggestions?

#include<stdio.h>

void show_it(int x,int y,int z)
{
printf("the id x is at addr: %8x\n",&x);
printf("the id y is at addr: %8x\n",&y);
printf("the id z is at addr: %8x\n",&z);
}

int main(void)
{
int x,y,z;

x=y=z=0;
show_it(x,y,z);
}

A run of it produces:

the id x is at addr: bffff9a0
the id y is at addr: bffff9a4
the id z is at addr: bffff9a8

so z is recognized first.

There are two aspects to this question

1. arguments are pushed onto the stack in right to left order, so
show_it(x,y,z);
pushes the arguments z, y, x
This is basically what the previous program tells you

2. arguments can be evaluated in any 'compiler' specific order, and this is something you can never find out (or rely on, even if you manage to figure it out).
show_it(x++,x++,x++);
is what you might try to figure this out, but any such programs are by definition undefined (read: wrong).

When a compiler evaluates the parameters and when it finally decides to push those arguments are different things.

Mmmm

Code:

---

#include <stdio.h>

int foo ( int x ) {
    printf( "%d\n", x );
    return x;
}

void show_it(int x,int y,int z) {
    printf("the id x is at addr: %p\n",&x);
    printf("the id y is at addr: %p\n",&y);
    printf("the id z is at addr: %p\n",&z);
}

int main ( ) {
    show_it( foo(1), foo(2), foo(3) );
    return 0;
}

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