C: Realloc twodimensional array created by Malloc

Question

I have a two-dimensional int array allocated as follows:

int **matrix = (int**)malloc(x * sizeof(int));
for (i = 0; i < x; i++)
    matrix[i] = (int*)malloc(y * sizeof(int));

If I do realloc this array by the following code

int **matrix = (int**)realloc(matrix, x * sizeof(int));
for (i = 0; i < x; i++)
    matrix[i] = (int*)malloc((y) * sizeof(int));

will there be any leftover "tails" of second dimension of the array and should I use free() function for them before reallocating, or they will go away themselves?

There is no 2D array, nor something that can be used as one. A pointer is not an array! — too honest for this site, Mar 19 '16 at 21:18
Some of your `sizeof` expressions are wrong. When allocating/reallocating `matrix`, you need `x*sizeof(int *)`. — Tom Karzes, Mar 19 '16 at 21:22
[Also, do not cast the return value of `malloc`](http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc) — Antti Haapala -- Слава Україні, Mar 19 '16 at 21:22
Well I can access elements of this "array" by using indexes as always (array[i][j]) and easily passing it to other functions. That is the only things I need for now. Edit: Thanks Weather Vane — 1valdis, Mar 19 '16 at 21:22
That is because `sizeof(int)` and `sizeof(int*)` are the same on your system. But they might not be. — Weather Vane, Mar 19 '16 at 21:23
And yes, all of the memory allocated in the first loop will obviously be lost when you grow the array, since you don't free it and you don't reallocate it. You have 2 choices: (1) free it first, or (2) call `realloc` instead of `malloc` when `i` is less then the old value of `x`. — Tom Karzes, Mar 19 '16 at 21:24
Are you trying to grow a matrix and keep existing data intact? — n. m. could be an AI, Mar 19 '16 at 21:43
`int **matrix = (int**)realloc(matrix, x * sizeof(int));` would be illegal if the new `mtarix` is in the same scope as the old , and a bad idea if in a new scope — M.M, Mar 20 '16 at 21:18

score 1 · Accepted Answer · answered Mar 19 '16 at 21:17

1

You will completely lose data in your matrix, so this is not really reallocing anything; you could just as well free and malloc the top-level array, getting possibly better performance. Furthermore your code would leak memory, because the 2nd level allocations are not freed.

answered Mar 19 '16 at 21:17

Antti Haapala -- Слава Україні

129,958
22
279
321

As I suspected. Thanks! – 1valdis Mar 19 '16 at 21:26

chqrlie · Answer 2 · 2016-03-20T21:12:18.683

Here is how you can resize the 2D matrix while keeping the current values intact, initializing the newly allocated cells to 0 and freeing any extra leftovers:

// matrix was allocated to x and y
// reallocate it to newx, newy
for (i = newx; i < x; i++) { /* free unneeded rows */
    free(matrix[i]);
}
matrix = realloc(matrix, sizeof(*matrix) * newx);
for (i = x; i < newx; i++) { /* initialize the new row pointers */
    matrix[i] = NULL;
}
for (i = 0; i < newx; i++) {
    matrix[i] = realloc(matrix[i], sizeof(*matrix[i]) * newy);
    for (int j = y; j < newy; j++)
        matrix[i][j] = 0;
}

C: Realloc twodimensional array created by Malloc

2 Answers2