C - 2D Array malloc/free - Segmentation fault - c

Question

I have a very simple snippet below for which I am trying to figure out the reason for getting a segmentation fault.
int main (int argc, char** argv)
{
const int size = 2;
char** test1 = NULL;
int index = 0;
test1=(char**)malloc(sizeof(char*) * size);
if (test1 != NULL)
{
for (index = 0; index < size ; index++)
{
test1[index]=(char*)malloc(sizeof(char));
test1[index]='a';
}
//Removing this block does not result in seg fault - start
for (index = 0 ; index < size ; index++)
{
free(test1[index]); //Seg. fault here
}
//Removing this block does not result in seg fault - end
free(test1);
}
return 0;
}
If I remove the block enclosed within start and end comment - I do not see seg fault. But I think that would result in a leak.
Any help is much appreciated.

Answer 1

I think you meant to dereference the test1[index]. Your code overwrites the address of the allocated memory with 'a', hence when it tries to free the memory it segs fault because 'a' is not a valid address.
test1[index]=(char*)malloc(sizeof(char));
*test1[index]='a';
This works as well
test1[index][0]='a';

Answer 2

You start out fine:
test1=(char**)malloc(sizeof(char*) * size);
if (test1 != NULL) {
Your loop is not:
for (index = 0; index < size ; index++) {
test1[index]=(char*)malloc(sizeof(char));
test1[index]='a';
}
First, you allocating only one byte for what is supposed to be a row of characters (since you have only one "size" variable, I assume you want your 2d array to be square: 2x2. So you need to multiply by size here like you did in the outer loop. You don't need "sizeof(char)", that's just a long-winded way of typing "1".
But worse than that, after allocating the row too short, you then throw that memory away by overwriting the pointer with a character (you should have gotten a compiler warning here). That's disaster waiting to happen and a memory leak.
What you really mean is:
for (index = 0; index < size ; index++) {
test1[index]=malloc(size);
test1[index][0]='a';
}