Using free() on pointers of two dimensional array with multiple malloc()?

Question

I have a function that allocates a two dimensional array within a function returning a pointer to the array. Creating the array requires an array of pointers each of which contains the address of a row of the two dimensional array.

How do I correctly free these two malloc() calls outside of this function once I am done with this array?

int** allocateMatrix(int rows, int cols)
{
    int* arr = malloc(rows*cols*sizeof(int));
    int** matrix = malloc(rows*sizeof(int*));
    int i;
    for(i=0; i<rows; i++)
    {
        matrix[i] = &(arr[i*cols]);
    }
    return matrix;
 }

The function is used like this:

int** 2d_arr = allocateMatrix(row,cols);

Thanks!

dbush · Accepted Answer · 2018-11-15T19:20:06.837

4

You can only pass to free what was received from malloc. So the number of calls to free must be the same as the number of calls to malloc.

The first row of 2d_arr, i.e. 2d_arr[0], contains &arr[0*cols] == &arr[0] == arr. So you want to free that and matrix itself:

free(2d_arr[0]);
free(2d_arr);

edited Nov 15 '18 at 19:20

answered Nov 15 '18 at 18:52

dbush

205,898
23
218
273

Is it ok when i do: free(2d_arr[0]); outside the func – Tim Weissenfels Nov 15 '18 at 19:09
@TimWeissenfels Yes, since `2d_arr` contains the same value as `matrix`, the effect is the same. – dbush Nov 15 '18 at 19:10
Thank you for your quick response! but what I do not understand is why I only have to free() the first index from matrix :) – Tim Weissenfels Nov 15 '18 at 19:16
@TimWeissenfels You can only pass to `free` what was passed to `malloc`. So two of each. – dbush Nov 15 '18 at 19:20
@TimWeissenfels Glad I could help. Feel free to [accept this answer](https://stackoverflow.com/help/accepted-answer) if you found it useful. – dbush Nov 15 '18 at 19:25

score 0 · Answer 2 · answered Nov 15 '18 at 18:54

This will work:

void freeMatrix( int **matrix )
{
    free( matrix[ 0 ] );
    free( matrix );
}

Because, for i == 0, this code

matrix[i] = &(arr[i*cols]);

sets matrix[ 0 ] to the address of arr[0], which is the same value returned from the first malloc call:

int* arr = malloc(rows*cols*sizeof(int));

It's clearer if you write matrix allocation like

int** allocateMatrix(int rows, int cols)
{
    int** matrix = malloc(rows*sizeof(int*));
    matrix[ 0 ] = malloc(rows*cols*sizeof(int));

    int i;
    for(i=1; i<rows; i++)
    {
            matrix[i] = &(arr[i*cols]);
    }
    return matrix;
 }

Note the change in malloc() order, the direct assignment to matrix[ 0 ], and the change in loop index to start at 1.

score 0 · Answer 3 · answered Nov 15 '18 at 19:41

I suggest that you use a single malloc() to allocate the entire memory area so as to make it easier to manage as a single pointer.

I believe the following modified version of your function should work.

int** allocateMatrix(int rows, int cols)
{
    size_t  sMemSize = rows*sizeof(int*) + rows*cols*sizeof(int);
    int** matrix = malloc(sMemSize);
    int* arr = (int *) (matrix + rows);
    int i;
    for(i=0; i<rows; i++)
    {
        matrix[i] = &(arr[i*cols]);
    }
    return matrix;
 }

Then you can just do free(2d_arr); where int** 2d_arr = allocateMatrix (nRows, nCols);.

Using free() on pointers of two dimensional array with multiple malloc()?

3 Answers3