If you're not using the C99 variable length arrays (it appears you are, so see below), the usual solution is to pass in a pointer to the first element, along with any indexes you want to use for accessing the elements.
Here's a piece of code that prints out a range of an array, similar to what you're trying to do with your sort
.
#include <stdio.h>
static void fn (int *arr, size_t start, size_t end) {
size_t idx;
for (idx = start; idx <= end; idx++) {
printf ("%d ", arr[idx]);
}
putchar ('\n');
}
int main (void) {
int my_array[] = {9, 8, 7, 6, 5, 4, 3, 2, 1, 0};
fn (my_array, 4, 6);
return 0;
}
This outputs elements four through six inclusive (zero-based), giving:
5 4 3
A couple of points to note.
Using my_array
in that function call to fn
automatically "decays" the array into a pointer to its first element. This actually happens under most (not all) circumstances when you use arrays, so you don't have to explicitly state &(my_array[0])
.
C already has a very good sort function built in to the standard library, called qsort
. In many cases, that's what you should be using (unless either you have a specific algorithm you want to use for sorting, or you're doing a homework/self-education exercise).
If you are using real VLAs, you should be aware that the [*]
construct is only valid in the function prototype, not in an actual definition of the function.
So, while:
void xyzzy(int, int[*]);
is valid, the following is not:
void xyzzy(int sz, int plugh[*]) { doSomething(); }
That's because, while you don't need the size parameter in the prototype, you do very much need it in the definition. And, since you have it, you should just use it:
void xyzzy(int sz, int plugh[sz]) { doSomething(); }
The gcc
compiler actually has a reasonably clear error message for this, far better than the "needs to be bounded in the function declaration" one you saw:
error: ‘[*]’ not allowed in other than function prototype scope