How do you pass an array to a function like this: void fooboo(char array[i]);

Question

I am trying to understand exactly what this code is trying to accomplish. The function median was given, but I added the main function and typedef/prototypes in efforts to comprehend what it does, by passing something into the function. However I can figure out what or how to pass something to it. I understand that the function is some kind of sort. What I really need to know is what is exactly being passed to the function? An array of N index?

Thank you for any guidance!

#include <stdio.h>
#include <stdlib.h>

typedef unsigned char pix_t;
pix_t median(pix_t window[N]);

int main() {

    pix_t window[] = { 4, 3, 2, 1 };
    pix_t output;
    output = median(window[N]);

}

pix_t median(pix_t window[N])
{
    pix_t t[N], z[N];
    int ii, k, stage;

    // copy locally
    for (ii = 0; ii<N; ii++) z[ii] = window[ii];

    for (stage = 1; stage <= N; stage++) {
        k = (stage % 2 == 1) ? 0 : 1;
        for (ii = k; ii<N - 1; ii++) {
            t[ii] = MIN(z[ii], z[ii + 1]);
            t[ii + 1] = MAX(z[ii], z[ii + 1]);
            z[ii] = t[ii];
            z[ii + 1] = t[ii + 1];
        }
    }

    return z[N / 2];
}

Answer 1

Given the function signature

pix_t median(pix_t window[N])

a call like

median(window[N]);

is wrong. The function expects an array of pix_t with at least N elements ^Note, whereas, you're passing only a single variable of type pix_t.

Morale of the story:: Whenever in confusion, check the data types

The function should be called with an array, something like

#define N 10                                    //any number

int main(void) {                               //note the change

    pix_t window[N] = { 4, 3, 2, 1 };
    pix_t output;
    output = median(window);                   //passing the array
}

should do.

---

Point to note: despite the array notation used in the function signature

 pix_t median(pix_t window[N]) { //....

inside the function, window is not an array. Quoting C11, chapter §6.7.6.3

A declaration of a parameter as ‘‘array of type’’ shall be adjusted to ‘‘qualified pointer to type’’, where the type qualifiers (if any) are those specified within the [ and ] of the array type derivation. [....]

---

Note:

What I really need to know is what is exactly being passed to the function? An array of N index?

The meaning of "at least N elements" refers to the a guarantee that the array has enough storage space to hold N elements up to position N-1, not that indices N, N+1, N+2, ... are valid/addressable.

You can read it as: "I have a guarantee of at least N cells of storage space so that I can store at most N elements in N-1 valid positions."

However, it's the programmer's responsibility to keep track of these details manually to avoid indexing into an invalid location and causing a segmentation fault; the environment will not do this automatically for you.