How to create and return dynamic array with function parameters

Question

I have a problem returning dynamic array pointer with function parameter. I get segfault

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

void createArray(int *ptr, int n)
{
    ptr = malloc(n * sizeof(int));
    for(int i = 1; i <= n; ++i)
    {
        *(ptr + (i - 1)) = i*i;
    }
}

int main() {
    int *array = NULL;
    int n = 5;
    createArray(array, n);
    for(int i = 0; i < n; ++i)
    {
        printf("%d", array[i]);
    }
    return 0;
}

I have to fill my array with i*i, when I is from 1 to n. I don't get any errors or warnings. Just message about segmentation fault. Process finished with exit code 139 (interrupted by signal 11: SIGSEGV)

Remember that C only passes function arguments by value. That means the value used in the call is copied into the functions local argument variable. So modifying the local argument variable itself will only modify this copy, not the original value used in the call. Please do research about emulating pass by reference in C. — Some programmer dude
– Some programmer dude, Commented Nov 16, 2022 at 16:35
On another note, why *(ptr + (i - 1)) instead of the easier to read ptr[i - 1]? — Some programmer dude
– Some programmer dude, Commented Nov 16, 2022 at 16:36
The problem is *ptr[i - 1], because it's the same as *(ptr[i - 1]) while you need (*ptr)[i - 1] . — Some programmer dude
– Some programmer dude, Commented Nov 16, 2022 at 16:45
Alternatively, you could define a int *createArray(int n) function and make it return a pointer to the allocated memory. Call it as array = createArray(n);. — Ian Abbott
– Ian Abbott, Commented Nov 16, 2022 at 16:54

Peter Irich · Accepted Answer · 2022-11-16 18:27:09Z

1

Memory must be allocate in the calling function, but not in called. This variant works:

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

void createArray(int *ptr, int n){
 int i;
    for(i = 1; i <= n; i++)    {
        *(ptr + (i - 1)) = i*i;
// fprintf(stdout,"%d %d\n", i, *(ptr + (i -1)));fflush(stdout);
    }
}

int main() {
    int i, n, *array = NULL;
 void *pvc;
    n = 5;
    array = (int *)malloc(n * sizeof(int));
    createArray(array, n);
    for(i = 0; i < n; i++)    {
        fprintf(stdout,"%d %d\n", i, array[i]);fflush(stdout);
    }
 pvc = (void *)array;
 free(pvc);
    return 0;
}

answered Nov 16, 2022 at 18:27

Peter Irich

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2 Comments

Kurbamit Over a year ago

It could be allocated in other function too. Thank you.

Peter Irich Over a year ago

Yes, I saw example with **ptr, but not tried it.

Kurbamit · Accepted Answer · 2022-11-24 21:53:05Z

0

You can change pointer through function parameters like this:

void createArray(int **ptr, int n)
{
    *ptr = malloc(n * sizeof(int));
    for(int i = 1; i <= n; ++i)
    {
        (*ptr)[i - 1] = i*i;
    }
}

int main() {
    int *array = NULL;
    int n = 5;
    createArray(&array, n);

Remember to call function like this: createArray(&array, n);

edited Nov 24, 2022 at 21:53

answered Nov 16, 2022 at 18:53

Kurbamit

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