How do i dynamically allocate memory to a 2D array in C avoiding the stack overflow issue?

Question

pascal = (int **) malloc(no_of_rows * sizeof(int));

for (i = 0; i < no_of_rows; i++)
{
    *(pascal + i) = (int*) malloc(col * sizeof(int));
}

Can someone tell me what's wrong in this code as I am beginner to this language. I keep on getting problem of Stack Overflow? What could be its possible reasons and how could it be avoided?

Answer 1

pascal=(int **)malloc(no_of_rows * sizeof(int));

should be

pascal=malloc(no_of_rows * sizeof(int*));

Notice the additional * I added. In general, you can write it better:

pascal=malloc(no_of_rows * sizeof *pascal);

Note that casting mallocs results is unnecessary in C.

Answer 2

You already got the right answer from Mr. Blue Moon. However, to clarify about the error in your case, I want to add an elaboration.

For the code

pascal = (int **) malloc(no_of_rows * sizeof(int));

it looks like, your pascal is of type int **, that is, pointer to a pointer-to-int.

So, while allocating a valid memory to pascal, you need to allocate the size equal to the element it points to, and it points to another integer pointer.

So, your allocation statement should be

pascal=malloc(no_of_rows * sizeof(int*));

Point to notice:

1. See why not to cast the return value of malloc() and family in C.
2. The allocation size is calculated as sizeof(int *) (size of a pointer to int) instead of sizeof(int).

Now, to make this statement more robust, we can (should) re-write it as

pascal=malloc(no_of_rows * sizeof*pascal);

Here, once again, two things to notice,

1. sizeof is an operator, it does not need the parenthesise around its operand unless it is a type name.
2. This statement, is independent of the type of pascal. even if you change the type of pascal, you don't need to change the above statement.

That said, always check for the success of malloc() before using the returned pointer.

Hope this helps.