Question

Using what I have learned here: How to use realloc in a function in C, I wrote this program.

int data_length; // Keeps track of length of the dynamic array.
int n; // Keeps track of the number of elements in dynamic array.

void add(int x, int data[], int** test)
{

    n++;

    if (n > data_length)
    {
        data_length++;
        *test = realloc(*test, data_length * sizeof (int));
    }

    data[n-1] = x;

}

int main(void)
{

    int *data = malloc(2 * sizeof *data);

    data_length = 2; // Set the initial values.
    n = 0;

    add(0,data,&data);
    add(1,data,&data);
    add(2,data,&data);

    return 0;
}

The goal of the program is to have a dynamic array data that I can keep adding values to. When I try to add a value to data, if it is full, the length of the array is increased by using realloc.

Question

This program compiles and does not crash when run. However, printing out data[0],data[1],data[2] gives 0,1,0. The number 2 was not added to the array.

Is this due to my wrong use of realloc?

Additional Info

This program will be used later on with a varying number of “add” and possibly a “remove” function. Also, I know realloc should be checked to see if it failed (is NULL) but that has been left out here for simplicity.

I am still learning and experimenting with C. Thanks for your patience.

Answer 1

Your problem is in your utilisation of data, because it points on the old array’s address. Then, when your call realloc, this area is freed. So you are trying to access to an invalid address on the next instruction: this leads to an undefined behavior.

Also you don’t need to use this data pointer. test is sufficient.

(*test)[n-1] = x;