pointer on 2d array in a void function

Question

Well, I would like to initiate my 2D array by a void function. But I obtain a Segmentation Fault...

That's my code :

 #include <stdio.h> #include <stdlib.h> #include <string.h> void groupeetchoixgen(int*** choixGEN); void main(int args, char **argv) { int** choixGEN; int i,j; choixGEN=(int**) malloc (sizeof(int*)*2); for (i=0; i<3; i++) { choixGEN[i]=(int*) malloc (sizeof(int)*3); } groupeetchoixgen(&choixGEN); } void groupeetchoixgen(int*** choixGEN) { (*(choixGEN)[1])[0]=1; }

I Think that the trouble is (*(choixGEN)[1])[0]=1; But I don't know why !

Thanks for your help

can you paste your error message here?

Hans Z
– Hans Z

2012-07-20 19:43:27 +00:00
Commented Jul 20, 2012 at 19:43 — Hans Z
– Hans Z, Commented Jul 20, 2012 at 19:43
shouldn't it just be (*choixGEN)[1][0] = 1; ?

stefan
– stefan

2012-07-20 19:44:25 +00:00
Commented Jul 20, 2012 at 19:44 — stefan
– stefan, Commented Jul 20, 2012 at 19:44

fferen · Accepted Answer · 2012-07-20 19:44:58Z

On this line:

 choixGEN=(int**) malloc (sizeof(int*)*2);

you are only allocating space for 2 int*s, but you access the 3rd element in the for loop.

Musa · Accepted Answer · 2012-07-20 19:52:52Z

1

You only allocate memory for two (int*) but you try to reference choixGEN[0],choixGEN[1],choixGEN[2] which is 3

edited Jul 20, 2012 at 19:52

answered Jul 20, 2012 at 19:44

Musa

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Comments

cybertextron · Accepted Answer · 2012-07-21 00:07:58Z

In General, SegFault in C means:

You're accessing a location in memory which was not allocated.
You're using a dangling pointer to access a memory location

So, most likely, your problem is in this line:

choixGEN=(int**) malloc (sizeof(int*)*2);

You declared a int*** choixGEN but only allocated memory for (int**)

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