C: How to pass a double pointer to a function

Question

I am getting an segmentation fault when I pass the double pointers to the function to initialize the memory

int main()
{
    double **A;
    initialize(A, 10, 10);
 ......
}

void initialize(double **A, int r, int c)
{
   A = (double **)malloc(sizeof(double *)*r);
   for(int i = 0; i< r; i++) {
        A[i] = (double *)malloc(sizeof(double) *c);
        for(int j = 0; j < c; j++) {
            A[i][j] = 0.0;
        }
   }
}

How can I pass the double pointers to the functions.....

Answer 1

Like others have said, you need to take a pointer to pointer to pointer in your init function. This is how the initialize function changes:

void initialize(double ***A, int r, int c)
{
   *A = (double **)malloc(sizeof(double *)*r);
   for(int i = 0; i< r; i++) {
        (*A)[i] = (double *)malloc(sizeof(double) *c);
        for(int j = 0; j < c; j++) {
            (*A)[i][j] = 0.0;
        }
   }
}

And main will be:

int main()
{
    double **A;
    initialize(&A, 10, 10);
}

Also, the code as you posted it should cause no segmentation fault when passing the A pointer in. The segmentation fault most likely occurs when you return from the function and try to access A, because the A in main will not have been initialized. Only a copy of it is initialized the way you do it, and that copy is local to the initialize function, so it's lost when you return.

Answer 2

If you want to modify a pointer to pointer you need to pass a pointer to pointer to pointer.

void func(double ***data) { *data = malloc(sizeof(double*)*10); for.... };
double ** data; func(&data);

Answer 3

Well for one thing, the A inside initialize is a copy of the A in main -- so when you get back to main, its A is still uninitialized. If you try and use it -- boom!

To pass it to initialize 'by reference', you need to change the parameter type to double*** and pass in &A in main. Then, when you use it in initialize, you need to dereference it each time, i.e. *A.