Question

In this toy code example:

int MAX = 5;

void fillArray(int** someArray, int* blah) {
  int i;
  for (i=0; i<MAX; i++)
    (*someArray)[i] = blah[i];  // segfault happens here
}

int main() {
  int someArray[MAX];
  int blah[] = {1, 2, 3, 4, 5};

  fillArray(&someArray, blah);

  return 0;
}

… I want to fill the array someArray, and have the changes persist outside the function.

This is part of a very large homework assignment, and this question addresses the issue without allowing me to copy the solution. I am given a function signature that accepts an int** as a parameter, and I’m supposed to code the logic to fill that array. I was under the impression that dereferencing &someArray within the fillArray() function would give me the required array (a pointer to the first element), and that using bracketed array element access on that array would give me the necessary position that needs to be assigned. However, I cannot figure out why I’m getting a segfault.

Many thanks!

Answer 1

I want to fill the array someArray, and have the changes persist outside the function.

Just pass the array to the function as it decays to a pointer to the first element:

void fillArray(int* someArray, int* blah) {
    int i;
    for (i=0; i<MAX; i++)
        someArray[i] = blah[i]; 
}

and invoked:

fillArray(someArray, blah);

The changes to the elements will be visible outside of the function.

If the actual code was to allocate an array within fillArray() then an int** would be required:

void fillArray(int** someArray, int* blah) {
    int i;
    *someArray = malloc(sizeof(int) * MAX);
    if (*someArray)
    {
        for (i=0; i<MAX; i++)  /* or memcpy() instead of loop */
            (*someArray)[i] = blah[i];
    }
}

and invoked:

int* someArray = NULL;
fillArray(&someArray, blah);
free(someArray);