Question

I have the following code:

int takeEven(int *nums, int numelements, int *newlist) {
    newlist = malloc(numelements * sizeof *newlist);
    int i, found = 0;
    for(i = 0; i < numelements; ++i, nums++) {
        if (!(*nums % 2)) {
            *(newlist++) = *nums;
            found++;
        }
    }
    newlist -= found;
    printf("First number found %d\n", *newlist); // <= works correctly
    return found;

}

int main()
{
    int nums[] = {1,2,3,4,5};
    int *evenNums;
    int i;
    int n = takeEven(nums, sizeof(nums) / sizeof(*nums), evenNums);
    for (i = 0; i < n; ++i) {
        printf("%d\n", *(evenNums++));
    }
    return 0;
}

The output of the above code:

-1
2088999640
2088857728

---

If I try printing the first element of the newlist pointer before returning the function (printf("First number found %d\n", *newlist);), it works as intended, but why is it that when I try to access the pointer from outside of the function I get those values from seemingly unmalloced addresses?

Answer 1

You need to pass in a pointer to pointer, i.e. int **newlist. Specifically, newlist is being passed into your function by value, so the newlist in main and inside your function are two completely different variables.

There is also a bug in your test for even numbers:

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

int takeEven(int *nums, int numelements, int **newlist) {
    int *list = malloc(numelements * sizeof **newlist);
    *newlist = list;  // this modifies the value of newlist in main
    int i, found = 0;
    for(i = 0; i < numelements; ++i, nums++) {
        if ((*nums % 2) == 0) {
            *(list++) = *nums;
            found++;
        }
    }
    list -= found;
    printf("First number found %d\n", *list); // <= works correctly
    return found;
}

int main()
{
    int nums[] = {1,2,3,4,5};
    int *evenNums;
    int i;
    int n = takeEven(nums, sizeof(nums) / sizeof(*nums), &evenNums);
    for (i = 0; i < n; ++i) {
        printf("%d\n", *(evenNums++));
    }
    return 0;
}

You can also take a look at this question from the C-FAQ which deals with your problem also:

Q: I have a function which accepts, and is supposed to initialize, a pointer:

void f(int *ip)
{
    static int dummy = 5;
    ip = &dummy;
}

But when I call it like this:

int *ip;
f(ip);

the pointer in the caller remains unchanged.

A: Are you sure the function initialized what you thought it did? Remember that arguments in C are passed by value. In the code above, the called function alters only the passed copy of the pointer. To make it work as you expect, one fix is to pass the address of the pointer (the function ends up accepting a pointer-to-a-pointer; in this case, we’re essentially simulating pass by reference):

void f(ipp)
int **ipp;
{
    static int dummy = 5;
    *ipp = &dummy;
}

...

int *ip;
f(&ip);

Another solution is to have the function return the pointer:

int *f()
{
    static int dummy = 5;
    return &dummy;
}

...

int *ip = f();

See also questions 4.9 and 4.11.