realloc 2D arrays in C

Question

I'm trying to increase a 2D array with malloc and realloc. I get a segmentation fault.

Here is how I create my tab:

char **tab2D = creerTab2D(taillex, tailley); 


char** creerTab2D(int taillex, int tailley) {

    char **tab2D = malloc( sizeof(char*) * taillex);

    for (int i = 0; i < taillex; i++) {
        tab2D[i] = creerTab1D(tailley);
    }

    return tab2D;
}

Here is how I increase his rows number by 1:

    (*taillex)++;

    *tab = (char**) realloc(*tab, *taillex * sizeof(char*)) ;

    if (*tab == NULL) { printf("Erreur lors du realloc\n"); return;}

    printf("%d", *taillex);

    *tab[*taillex] = malloc(*tailley * sizeof(char)); // Here is the bad access (segmentation fault)

    for (int i = 0; i < *tailley; i++) {
        (*tab)[*taillex][i] = '*';
        printf("%c", (*tab)[*taillex][i]);
    }

This piece of code is inside a function with this prototype (too big to paste here):

void insererCharDansTab2D(char ***tab, int *taillex, int *tailley, int posx, int posy, char c)

I wrote a comment where the segmentation fault appears. Thanks mates!

_sigh_ Thereis no 2D array in your code. A pointer is not an array! And being a 3-star C programmer is not a compliment. — too honest for this site, Apr 04 '17 at 16:10
`*tab[*taillex] = malloc(*tailley * sizeof(char));` --> `(*tab)[*taillex - 1] = malloc(*tailley * sizeof(char));` — BLUEPIXY, Apr 04 '17 at 16:13
@BLUEPIXY Thanks a lot that works! I'm stupid ahah that was so evident... — Paul Bénéteau, Apr 04 '17 at 16:19
You cannot realloc a 2D array without risking corruption of the existing data. — ThingyWotsit, Apr 04 '17 at 16:20
@ThingyWotsit ???? I learned that the computer will try to find another place in the memory to put the new array. — Paul Bénéteau, Apr 04 '17 at 16:23
@greenpoisononeTV An up-vote if you can show me the 2D array in this post's code. I see pointers to pointers and arrays of pointers — chux - Reinstate Monica, Apr 04 '17 at 16:25
@greenpoisononeTV think about it. After the realloc/copy, all the original data will be there, but not in the places you expect... — ThingyWotsit, Apr 04 '17 at 16:27
I you have a 1d array of pointers that points each other to other 1d arrays... — Paul Bénéteau, Apr 04 '17 at 16:28
@greenpoisononeTV Exactly... so you have a lot of 1D arrays, not a 2D array. — Dmitri, Apr 04 '17 at 16:29
The solution gived by @BLUEPIXY is only working for arrays with a size greater than 10... — Paul Bénéteau, Apr 04 '17 at 16:29
@greenpoisononeTV: But that's not a 2D *array*. In an array, all elements are *contiguous*. An array of pointers to other arrays is not *contiguous*. — John Bode, Apr 04 '17 at 16:30
@greenpoisononeTV: Yes, you can *index* it like a 2D array, but if you need to treat it as a stream of bytes (say for serialization purposes), then it's not the same thing as a 2D array. — John Bode, Apr 04 '17 at 16:31
Example: `char a2d[3][4]` is a [2D array](https://cdecl.org/?q=char+a2d%5B3%5D%5B4%5D). `char **ptr2` is a [pointer to pointer to char](https://cdecl.org/?q=char+a2d%5B3%5D%5B4%5D) — chux - Reinstate Monica, Apr 04 '17 at 16:32
@chux yeah but i can get a value like this : `(*tab)[*taillex][i]`. Is it not the same? — Paul Bénéteau, Apr 04 '17 at 16:34
@greenpoisononeTV `8.0/2.0` and `8/2` both have the same _value_ of 4, but are not the same type. — chux - Reinstate Monica, Apr 04 '17 at 16:36
With a 2D array, `&arr[0][0]`, `&arr[0]`, and the decay of `arr` will all be the same address (but different pointer types). With what you have, they'll be different because your version has several 1D arrays in different areas of memory... your version also has to store the individual row addresses in another array, which isn't needed for a real 2D array. — Dmitri, Apr 04 '17 at 16:37
Did you look at the duplicate suggested? It has an answer that should solve your issue. — Dmitri, Apr 04 '17 at 16:44
OK I found from mylself that's good.. Issue solved! Thanks for all mates! ;) — Paul Bénéteau, Apr 04 '17 at 16:45

realloc 2D arrays in C

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