Double pointer doesn't get malloc'd in function

Question

So I'm working in C and I have an int ** inside of one function that is modified within another function. I was getting a SegFault when I was running this problem so I decided to debug it with gdb . What I found was that memory was never allocated to this array. My functions are like this

void declarer()
{
    int ** a;
    alocator(a, 4, 5);
    for (int i = 0; i < 4; i++)
        for (int j = 0; j < 5; j++)
            printf("%d\n", a[i][j]);
}

void alocator(int ** a, int b, int c)
{
    a = (int **)malloc(sizeof(int *) * b);
    for (int i = 0; i < b; i++) {
        a[i] = (int *)malloc(sizeof(int) * c);
        for (int j = 0; j < c; j++)
            a[i][j] = j;
    }
}

When I run gdb with a breakpoint after the line alocator(a, 4, 5) (before the program segfaults), and I write pa , I get $1 = (int **) 0x0 which shows that a is at address 0x0 and has no memory allocated for it. Why did alocator not allocate memory to it, and how can I get alocator to allocate memory to a ?

Answer 1

Because you are passing **a by value, you have to pass it as pointer. So use a triple pointer:

void declarer()
{
    int **a;
    alocator(&a, 4, 5);
    ...
}

void alocator(int ***a, int b, int c)
{
    *a = (int **)malloc(sizeof(int *) * b);
    ...

Answer 2

You pass the variable a by value to the function, so its value won't be changed. You have two solutions:

Either change your function to accept value of type int *** and pass a by address, or change the return value of the function to int ** and make use of that value upon invocation.

example:

void declarer()
{
    int ** a = alocator(4, 5);
    for (int i = 0; i < 4; i++)
        for (int j = 0; j < 5; j++)
            printf("%d\n", a[i][j]);
}

int **alocator(int b, int c)
{
    int **a = (int **)malloc(sizeof(int *) * b);
    for (int i = 0; i < b; i++) {
        a[i] = (int *)malloc(sizeof(int) * c);
        for (int j = 0; j < c; j++)
            a[i][j] = j;
    }
    return a;
}