C Pointer Assignment in a function

Question

Why can't I assign a point in a function. As you noticed in the following code. I can not assign pointer p1 pointed to the correct address after the function return. But with the global pointer *p, I can store the address information.

#include <stdio.h>
#include <stdlib.h>

int *p = NULL;
void test(int * pt1, int**pt2){
    p = (int*)malloc(sizeof(int));    
    pt1 = p;
    *pt2 = p;
    printf("p points to %p\n", p);
    printf("pt1 points to %p\n", pt1);
    printf("pt2 points to %p\n", *pt2);
}

int main(void) {
    int *p1 = NULL; 
    int *p2 = NULL;

    printf("p points to %p\n", p);
    printf("p1 points to %p\n", p1);
    printf("p2 points to %p\n", p2);

    test(p1, &p2);

    printf("p points to %p\n", p);
    printf("p1 points to %p\n", p1);
    printf("p2 points to %p\n", p2);

    return 0;
}

OUTPUT:

p points to (nil)
p1 points to (nil)
p2 points to (nil)
p points to 0x8acb008
pt1 points to 0x8acb008
pt2 points to 0x8acb008
p points to 0x8acb008
p1 points to (nil)
p2 points to 0x8acb008

In C, everything is passed by value. Pointers ain't no exception. — user529758
– user529758, Commented Feb 7, 2013 at 20:27

Flexo - Save the data dump · Accepted Answer · 2013-02-07 20:40:03Z

Inside test the variable pt1 is a discrete pointer in its own right. That is to say it is not merely an alias for p1, but rather a copy that exists only for the lifetime of the call.

Thus whatever assignment you make to it is only exits for the duration of that call and is not propagated outside of it. When you return from test the pointer pt1 ceases to exist and any changes aren't copied back.

As well as using an extra "layer" of pointer like you did with pt2 sometimes it's appropriate to use the return value to "share" changes with a wider audience:

#include <stdio.h>
#include <stdlib.h>

int *p = NULL;
int *test(int * pt1, int**pt2){
    p = (int*)malloc(sizeof(int));    
    pt1 = p;
    *pt2 = p;
    printf("p points to %p\n", p);
    printf("pt1 points to %p\n", pt1);
    printf("pt2 points to %p\n", *pt2);
    return pt1;
}

int main(void) {
    int *p1 = NULL; 
    int *p2 = NULL;

    printf("p points to %p\n", p);
    printf("p1 points to %p\n", p1);
    printf("p2 points to %p\n", p2);

    p1=test(p1, &p2);

    printf("p points to %p\n", p);
    printf("p1 points to %p\n", p1);
    printf("p2 points to %p\n", p2);

    return 0;
}

75inchpianist · Accepted Answer · 2013-02-07 20:29:24Z

You arre passing the p1 in by value, so the changes are only visable within the scope of that function. pass in a pointer to that pointer, as you did with p2, and you are fine.

#include <stdio.h>
#include <stdlib.h>

int *p = NULL;
void test(int **pt1, int**pt2){
    p = (int*)malloc(sizeof(int));    
    *pt1 = p;
    *pt2 = p;
    printf("p points to %p\n", p);
    printf("pt1 points to %p\n", pt1);
    printf("pt2 points to %p\n", *pt2);
}

int main(void) {
    int *p1 = NULL; 
    int *p2 = NULL;

    printf("p points to %p\n", p);
    printf("p1 points to %p\n", p1);
    printf("p2 points to %p\n", p2);

    test(&p1, &p2);

    printf("p points to %p\n", p);
    printf("p1 points to %p\n", p1);
    printf("p2 points to %p\n", p2);

    return 0;
}

William · Accepted Answer · 2013-02-07 20:30:04Z

1

You're passing p1 by value, so it's not updated in the main function. However, you are passing p2 by reference (note that you wrote &p2), so it can be changed.

answered Feb 7, 2013 at 20:30

William

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1 Comment

William Over a year ago

Oops, looks like 75inchpianist beat me to it!

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