Variable Sized Arrays in C

Question

On the discussion of dynamic memory here: "Intro to C Pointers and Dynamic Memory"

The author states:

A memory block like this can effectively be used as a more flexible array. This approach is actually much more common in real-world C programs. It's also more predictable and flexible than a "variable size array"

The type of memory block he is talking about is as such:

const int size = 5;
int * array = calloc(size, sizeof(int));

and then using another pointer to iterate over the array:

int * index = array;
for (i = 0; i < size; i++) {
    *index = 1; // or whatever value
    index++;
}

My question is how is this method better than a standard variable sized array like this?:

int array[variable];

or dynamic:

char name[] = "Nick";

The author doesn't really shed much light as to why I should prefer the former method to the latter. Or more specifically: How is it more "predictable and flexible"?

Answer 1

If you declare int array[variable] the memory will be allocated on the stack which is not very good for large, relatively permanent data structures (such as one you might want to return). You don't need to free memory manually if you use the array syntax since it's freed when it goes out of scope. calloc on the other hand will allocate memory dynamically at run time on the heap. You'll have to free it yourself as soon as you're finished with it.

Answer 2

Because

int array[variable];

isn't valid in standard C -- you can only define the length of an array with a constant. (such as your

char name[] = "Nick";

example, which isn't variable-length).

As such, it's necessary to use a memory allocator like calloc() if you want to create an array of a length based on a program variable.

Just don't forget to free() it.

Answer 3

I agree with ocdecio that c89 doesn't allow

int array[variable]

c99 allows some types of variables and expressions to be the array size. But in addition to that, things allocated with malloc and family can be resized using realloc.