Passing dynamically allocated 2d array to a function - c

Question

The following code is not working correctly. I'm getting a segfault when I run the program. I ran my program through gdb and found out that the error is occuring in the fillArrays(int**,int) function.
GDB is displaying the following parameters for fillArrays(int**,int):
fillArrays (arrays=0x0,numArrays=3)
Here is the source code to my program
#include <stdlib.h> /* malloc and free */
#define MULTIPLIER 1
#define SMALL 10
#define BIG 20
void allocateSmallArrays(int **arrays,int numArrays) {
int index,freeIndex;
int outerIndex,innerIndex;
arrays = malloc(numArrays*sizeof(int*));
if(arrays == NULL) {
printf("out of memory\n");
exit(1);
}
for(index = 0;index < numArrays;index++) {
arrays[index] = malloc(SMALL*sizeof(int));
if(arrays[index] == NULL) {
printf("out of memory\n");
exit(1);
}
}
}
void fillArrays(int **arrays,int numArrays) {
int outerIndex,innerIndex;
for(outerIndex = 0;outerIndex < numArrays;outerIndex++) {
for(innerIndex = 0;innerIndex < SMALL;innerIndex++)
arrays[outerIndex][innerIndex] = 0;
}
}
void deallocateSmallArrays(int **arrays,int numArrays) {
int index;
for(index = 0;index < numArrays;index++)
free(arrays[index]);
free(arrays);
}
int main(void) {
int numArrays = (3 * MULTIPLIER);
int **arrays = 0;
allocateSmallArrays(arrays,numArrays);
fillArrays(arrays,numArrays);
deallocateSmallArrays(arrays,numArrays);
arrays = 0;
return 0;
}
I was under the assumption that since arrays was allocated in allocateSmallArrays, that passing it through fillArrays would 0 out the allocated arrays and then deallocate in the last function. How do I go about accomplishing this?

Answer 1

The problem is that allocateSmallArrays changes its own copy of the arrays pointer. So the result of the malloc is lost and after the function is done, in the caller arrays is still 0. You could:
Pass a triple pointer int ***arrays and do to *arrays everything you're doing to arrays
Return the pointer instead of void
A C FAQ deals with this very subject.