Question

I have written a some C code running on OS X 10.6, which happens to be slow so I am using valgrind to check for memory leaks etc. One of the things I have noticed whilst doing this:

If I allocate memory to a 2D array like this:

double** matrix = NULL;
allocate2D(matrix, 2, 2);

void allocate2D(double** matrix, int nrows, int ncols) {
    matrix = (double**)malloc(nrows*sizeof(double*));
    int i;
    for(i=0;i<nrows;i++) {
        matrix[i] = (double*)malloc(ncols*sizeof(double));
    }
}

Then check the memory address of matrix it is 0x0.

However if I do

double** matrix = allocate2D(2,2);

double** allocate2D(int nrows, int ncols) {
    double** matrix = (double**)malloc(nrows*sizeof(double*));
    int i;
    for(i=0;i<nrows;i++) {
        matrix[i] = (double*)malloc(ncols*sizeof(double));
    }
return matrix;
}

This works fine, i.e. the pointer to the newly created memory is returned.

When I also have a free2D function to free up the memory. It doesn’t seem to free properly. I.e. the pointer still point to same address as before call to free, not 0x0 (which I thought might be default).

void free2D(double** matrix, int nrows) {
    int i;
    for(i=0;i<nrows;i++) {
        free(matrix[i]);
    }
free(matrix);
}

My question is: Am I misunderstanding how malloc/free work? Otherwise can someone suggest whats going on?

Alex

Answer 1

When you free a pointer, the value of the pointer does not change, you will have to explicitly set it to 0 if you want it to be null.