Passing an uninitialized array to a function in C

Question

Purpose of my program is simple, to read sequence of int numbers from file, here's the code:

int main()
{
   FILE *file = fopen("test.txt", "r");

   int *array = NULL; // Declaration of pointer to future array;
   int count;

   process_file(file, array, &count);

   return 0;
}

// Function returns 'zero' if everything goes successful plus array and count of elements as arguments
int process_file(FILE *file, int *arr, int *count)
{
    int a;
    int i;
    *count = 0;

    // Function counts elements of sequence
    while (fscanf(file, "%d", &a) == 1)
    {
        *count += 1;
    }

    arr = (int*) malloc(*count * sizeof(int)); // Here program allocates some memory for array

    rewind(file);

    i = 0;
    while (fscanf(file, "%d", &a) == 1)
    {
        arr[i] =  a;
        i++;
    }

    fclose(file);

    return 0;
}

Problem is, in outer function (main), array haven't changed. How could it be fixed?

score 4 · Accepted Answer · edited May 23 '17 at 11:44

4

You need to pass the array by reference, so that the function can change its caller's array.

It must be:

process_file(FILE *file, int **arr, int *count)

and call like so:

process_file(file, &array, &count);

Additionally, I would suggest:

Using size_t rather than int for the array length.
Not casting the return value of malloc().
Computing size like so: *array = malloc(*count * sizeof *array);, this avoids repeating int.

edited May 23 '17 at 11:44

Community

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answered Jul 30 '16 at 10:28

unwind

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Shouldn't that be "*array = malloc(*count * sizeof **array);"? – Rudy Velthuis Jul 30 '16 at 12:35
thanks, it works with some adjustment: to assign value to array's element in function i need to make it like this (*array)[i] = value – jl284 Jul 30 '16 at 16:15

Passing an uninitialized array to a function in C

1 Answers1