Thread: Pointers: Call by value

Originally Posted by quzah

Probably every time you ever have a question. You see, it wasn't me who skipped out on your classes and didn't pay attention. That was you. As such, I see it as a good idea to make an example of you for all the other would-be dumbasses out there. Perhaps they'll learn from your mistakes.

Quzah.

Fair enough.

I won't be asking beginner questions forever though.

Originally Posted by quzah

Code:

#include <stdio.h>

void foo( int *a )
{
        printf("a is    : %p\n", (void *)a );
        printf("&a is   : %p\n", (void *)&a );
        printf("&a[0] is: %p\n", (void *)&a[0] );
}

void bar( int a[] )
{
        printf("a is    : %p\n", (void *)a );
        printf("&a is   : %p\n", (void *)&a );
        printf("&a[0] is: %p\n", (void *)&a[0] );
}

int main ( void )
{
        int a[5] = {0}, *b = a;

        printf("In main:\n");
        printf("a is    : %p\n", (void *)a );
        printf("&a is   : %p\n", (void *)&a );
        printf("&a[0] is: %p\n", (void *)&a[0] );
        printf("\n");

        printf("In foo(a):\n");
        foo( a );
        printf("\n");

        printf("In bar(a):\n");
        bar( a );
        printf("\n");

        printf("In foo(b):\n");
        foo( b );
        printf("\n");

        printf("In bar(b):\n");
        bar( b );
        printf("\n");

        return 0;
}


In main:
a is    : 0xbffffce0
&a is   : 0xbffffce0
&a[0] is: 0xbffffce0

In foo(a):
a is    : 0xbffffce0
&a is   : 0xbffffcc0
&a[0] is: 0xbffffce0

In bar(a):
a is    : 0xbffffce0
&a is   : 0xbffffcc0
&a[0] is: 0xbffffce0

In foo(b):
a is    : 0xbffffce0
&a is   : 0xbffffcc0
&a[0] is: 0xbffffce0

In bar(b):
a is    : 0xbffffce0
&a is   : 0xbffffcc0
&a[0] is: 0xbffffce0

The name of an array is in effect a pointer to its first element. You'll see that in every example here. What changes is the address of the array or pointer itself. When passed to a function, "array[]" is changed to "*array", or, a pointer to that type. This is illustrated by the functions foo and bar.

The only thing different in the output is as it should be: The address of the pointer itself is different then what you get inside main. The reason is, it's a different pointer. If you'll recall, if you can, a pointer is a variable that stores an address. Thus, the pointer itself has an address. Since it is an actual different variable, it has a (shock!) a different address. This you'll see as the second line in the output of functions foo and bar.

Quzah.

Thank you for giving the code and explanation.

I understand the second portion of the post, that the reason for a different address is because it's a different variable. But why is it the same in the main?

You said the name of an array is in effect a pointer to its first element, so it's already a pointer in main. So why is the address of the pointer, holding the same address of it's variable/address it points to? Does it point to itself?

The address changes in the function because the pointer that is the array's name is passed by value, but why it's the same in main I don't understand.

Thanks again, from a dumbass.

The FAQ was excellent. Thanks.