How to malloc an array in a function, then realloc it

Question

I want to pass an array pointer to a function, then malloc and realloc it. The function end, but after I get a Segmentation fault (core dumped) error before returning to the main.

I'm not too familiar with C pointers, I used mainly C++ for my codes, don't know how to use pointers as references.

Main:

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

int main(){
int A[]={1,2,4,5,7};
int B[]={1,2,3,4,6,8};
int *C=NULL;
int k=0;
egyesit(A,B,C,5,6,&k);
int i;
for (i = 0; i < k; i++)
    {
        printf("%d ", C[i]);
    }
    printf("\n");
return 0;

Header:

void egyesit(int *A, int *B, int *C, int n,int m,int *k);

Function in a different c file:

#include <stdio.h>
#include <stdlib.h>
void egyesit(int *A, int *B, int *C, int n, int m, int *k)
{
    int i, j;
    C = (int *)malloc(n * sizeof(int));
    *k = n;
    for (i = 0; i < n; i++)
    {
        printf("%d ", A[i]);
        C[i] = A[i];
    }
    printf("\n");

    for (i = 0; i < *k; i++)
    {
        printf("%d ", C[i]);
    }
    printf("\n");

    int ok;
    for (i = 0; i < m; i++)
    {
        ok = 1;
        printf("%d ", B[i]);
        for (j = 0; j < *k; j++)
        {
            if (C[j] == B[i])
            {
                ok = 0;
                break;
            }
        }
        if (ok == 1)
        {
            *k=*k+1;
            C = (int *)realloc(C, *k * sizeof(int));
            C[*k - 1] = B[i];
        }
    }
    printf("\n");
    for (i = 0; i < *k; i++)
    {
        printf("%d ", C[i]);
    }
    printf("\n");
}

`egyesit(A,B,C,5,6,&k);` gives a copy of the _pointer_ `C` to `egyesit()`. Changing `C` within `egyesit()` (as in `C = (int *)malloc(n * sizeof(int));`) will not affect the `C` in the calling code. Consider passing the _address_ of `C`, just like code pass in the address of `k`. — chux - Reinstate Monica, May 11 '18 at 18:04
Ok, how can I modify C from the calling code? I need to use **C, and &C in the main? — zarnilord, May 11 '18 at 18:07
Looks about right. Try it and see. Be sure to have all compiler warnings enabled to save time. — chux - Reinstate Monica, May 11 '18 at 18:08

score 1 · Accepted Answer · answered May 11 '18 at 18:10

To allocate malloc or realloc momory for C you have to pass its address to the function. So your function should be:

void egyesit(int *A, int *B, int **C, int n,int m,int *k)

otherwise the pointer allocated in egyesit will not be set in the variableC declared in main.

So your function becomes like this:

void egyesit(int *A, int *B, int **C, int n, int m, int *k)
{
    int i, j;
    *C = (int *)malloc(n * sizeof(int));
    *k = n;
    for (i = 0; i < n; i++) {
        printf("%d ", A[i]);
        (*C)[i] = A[i];
    }
    printf("\n");

    for (i = 0; i < *k; i++) {
        printf("%d ", (*C)[i]);
    }
    printf("\n");

    int ok;
    for (i = 0; i < m; i++) {
        ok = 1;
        printf("%d ", B[i]);
        for (j = 0; j < *k; j++) {
            if ((*C)[j] == B[i]) {
                ok = 0;
                break;
            }
        }
        if (ok == 1) {
            *k = *k + 1;
            *C = (int *)realloc(*C, *k * sizeof(int));
            (*C)[*k - 1] = B[i];
        }
    }
    printf("\n");
    for (i = 0; i < *k; i++) {
        printf("%d ", *C[i]);
    }
    printf("\n");
}

And then you call it from main in this way:

egyesit(A, B, &C, 5, 6, &k);

score 0 · Answer 2 · answered May 11 '18 at 18:17

Or you could just return a pointer to the memory allocated in the function eyesgit() as in

int *egyesit(int *A, int *B, int *C, int n, int m, int *k)
{
    int i, j;
    C = malloc(n * sizeof(int));

    ....
    ....
    ....


    return C;
}

and in main(),

C=egyesit(A,B,C,5,6,&k);

But don't forget to deallocate the memory once you are done using it like

free(C);

You need not explicitly cast the return value of malloc(). Read this post.

malloc() and realloc() returns NULL if there some error like insufficient memory. You may want to check that as well.

score 0 · Answer 3 · answered May 11 '18 at 18:26

Segmentation fault is in your function egyesit in the first for loop

*C[i] = A[i];

This is because you allocated variable C like this

*C = (int *)malloc(n * sizeof(int));

C is pointer on pointer on int... int **C

So you should first allocate memory like this

C =(int**)malloc(sizeof(int*));

After that, you allocate memory for the array of type int on which C is pointing

//*C meaning you are dereferencig on what C is pointing, and allocate memory for that
*C=(int*)malloc(sizeof(int)*n));

This is how it should look like

 void egyesit(int *A, int *B, int *C, int n, int m, int *k)
 {
     int i, j;
     C = (int **)malloc(sizeof(int*));
     *C = (int*)malloc(sizeof(int)*n);
     *k = n;
     for (i = 0; i < n; i++)
     {
         printf("%d ", A[i]);
         C[i] = A[i];
     }
     printf("\n");
     //and the rest of the function

How to malloc an array in a function, then realloc it

3 Answers3