Executive summary:

int a[17];
size_t n = sizeof(a)/sizeof(a[0]);

Full answer:

To determine the size of your array in bytes, you can use the sizeof operator:

int a[17];
size_t n = sizeof(a);

On my computer, ints are 4 bytes long, so n is 68.

To determine the number of elements in the array, we can divide the total size of the array by the size of the array element. You could do this with the type, like this:

int a[17];
size_t n = sizeof(a) / sizeof(int);

and get the proper answer (68 / 4 = 17), but if the type of a changed you would have a nasty bug if you forgot to change the sizeof(int) as well.

So the preferred divisor is sizeof(a[0]) or the equivalent sizeof(*a), the size of the first element of the array.

int a[17];
size_t n = sizeof(a) / sizeof(a[0]);

Another advantage is that you can now easily parameterize the array name in a macro and get:

#define NELEMS(x)  (sizeof(x) / sizeof((x)[0]))

int a[17];
size_t n = NELEMS(a);

If its on the stack, you can use the sizeof operator to determine its raw byte size:

int array[5];
size_t array_size = sizeof(array);

But this gives you the size in bytes, not the number of elements. However, you can calculate the number of elements with this approach:

int array[5];
size_t array_elems = sizeof(array) / sizeof(* array);

If you array is a pointer (e.g. function call or dynamic memory allocation) then you have to keep track it yourself!