free multidimension char** in C not working

Question

I have this functions:

char** init_matrix(int rows, int columns){
    char **matrix = (char **)malloc (rows*sizeof(char *));
    for(int i = 0; i < rows; i++){
        matrix[i] = (char *) malloc (columns*sizeof(char));
        for(int j = 0; j < columns; j++){
            matrix[i][j] = '-';
        }
    }
    return matrix;
}

void show_matrix(char **matrix, int rows, int columns){
    printf("\n\n");
    for(int i = 0; i < rows; i++) {
        for(int j = 0; j < columns; j++)
           printf("|%c|", matrix[i][j]);
        printf("\n");
        for(int j = 0; j < columns; j++)
            printf("---");
        printf("\n");
    }
}

void setValueInMatrix(char** matrix, int row, int column, char value){
    matrix[row][column] = value;
}

Then i do this

char **matrix = init_matrix(rows, columns);
setValueInMatrix(matrix, solucion->row, solucion->column, solucion->aminoacido);
printf("matrix before free\n");
show_matrix(matrix, rows, columns);
    for(int i = 0; i < rows; i++){
        free(matrix[i]);
    }
    free(matrix);
printf("matrix after free\n");
show_matrix(matrix, rows, columns);

Why the output is the same before and after the free?? :S

Output:

matrix before free


|-||-||-||H||-||-||-|
---------------------
|-||-||-||P||-||-||-|
---------------------
|-||-||-||P||-||-||-|
---------------------
|-||-||-||H||-||-||-|
---------------------
|-||-||-||-||-||-||-|
---------------------
|-||-||-||-||-||-||-|
---------------------
|-||-||-||-||-||-||-|
---------------------
matrix after free


|-||-||-||H||-||-||-|
---------------------
|-||-||-||P||-||-||-|
---------------------
|-||-||-||P||-||-||-|
---------------------
|-||-||-||H||-||-||-|
---------------------
|-||-||-||-||-||-||-|
---------------------
|-||-||-||-||-||-||-|
---------------------
|-||-||-||-||-||-||-|
---------------------

I have read all this posts but it doesnt look like it works:

`show_matrix(matrix, rows, columns);` after you've free'd the content it inspects therein invokes *undefined behavior*. In short, what you see from that point on *is irrelevant*, as your expectations for definitive behavior have been squelched by your own hand. — WhozCraig, Jun 21 '16 at 18:34
`char **` is a scalar. It has no dimension(s)! If you mean 2D array: `char **` is no 2D array and cannot represent one. — too honest for this site, Jun 21 '16 at 18:57
I wasnt expectating any output, just my program was eating all RAM in seconds, now it works fine. — Victor Company, Jun 21 '16 at 19:11

Cherubim · Accepted Answer · 2016-06-21T20:14:49.103

3

Why the output is the same before and after the free

This is undefined behaviour
Once you've freed memory you must remember not to use it any more.
After calling
```
free(matrix);
```

you are getting the same output because in your case

the matrix still points at the same memory.

However, since matrix is freed, it's now available to be used again

Note: the matrix might or might not point the same block after freeing, it's undefined behavior but, in your case it does point the same block

know more here : click

To avoid this assign a pointer to NULL whenever you free it. This is a good practice.

matrix=NULL;

edited Jun 21 '16 at 20:14

answered Jun 21 '16 at 18:24

Cherubim

5,287
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`you are getting the same output because the matrix still points at the same memory.` __NO__, that's not guranteed. Only this holds true here is UB. – Sourav Ghosh Jun 21 '16 at 19:57
Yes the pointer may or not point the same memory... But it does in this case @SouravGhosh – Cherubim Jun 21 '16 at 20:01
... which is another outcome of UB. – Sourav Ghosh Jun 21 '16 at 20:03
1

Fine @SouravGhosh I'll edit my post – Cherubim Jun 21 '16 at 20:04
1

Sure, thanks for understanding. Cheers!! – Sourav Ghosh Jun 21 '16 at 20:06

score 2 · Answer 2 · answered Jun 21 '16 at 18:23

The output is the same because free marks the memory as free (available for reuse). It does not physically erase the memory.

But using a pointer after you have freed it is something the code should not do because you are reading/writing memory that may have been allocated for a different purpose.

Once you "free" memory you have to consider the pointer variable (matrix) as invalid. Similarly calling free() on the same pointer twice is known as a double-free and will crash your program or lead to other bad behavior.

Many people set the pointer variable to NULL after they free it to prevent accidental reuse.

Like many things in 'C' the language specification allows you to violate the rules, but you do so at your peril.

The language specification does not allow you to violate the rules; your compiler does. — Lightness Races in Orbit, Jun 21 '16 at 18:33
@LightnessRacesinOrbit and to add, just because you _can_ shoot your foot, does not mean you _have to_. :) — Sourav Ghosh, Jun 21 '16 at 18:54

free multidimension char** in C not working

2 Answers2