Question

Before variable-length arrays were supported, I would dynamically allocate them like this:

int foo(size_t n)
{
    int *arr = malloc(n * sizeof int);
    if (!arr) return ENOMEM; /* not enough memory */
    .
    . else do stuff with arr[]
    .
    free(arr);
    return 0;
}

With variable-length arrays I can now make it look cleaner:

int bar(size_t n)
{
    int arr[n];
    .
    . do stuff with arr[]
    .
    return 0;
}

But now I have no “out of memory” checking. In fact, the program crashes if n is too big.

How can I gracefully bail from bar(n) if n is too big?

Answer 1

The situation is exactly unchanged from any other local variables – a declaration like this:

int baz(void)
{
    int arr[100000];
    .
    . do stuff with arr[]
    .
    return 0;
}

has exactly the same problem. The “solution” is the same as it always has been – don’t recurse too deeply, and don’t allocate very large data structures with automatic storage duration (continue to use malloc() for these cases). The value of “very large” depends strongly upon your environment.

In other words, don’t declare int array[n]; unless you know that n is bounded to a reasonable value, such that you would have been happy to declare an array of that maximum size as an ordinary, non-variably-modified type array.

(Yes, this means that variably-modified type arrays are not as useful as they first appear, since you gain very little over just declaring the array at the maximum needed size).