Passing an array into a function, nothing outputting - c++

Question

I am trying to create something that fills in an array with 100 random numbers between 1 and 100. This was working fine when it was in the main function, but when I put it in an int function nothing outputs; I must be missing something simple as I'm just beginning. What can I do to fix this?
#include "stdafx.h"
#include <time.h>
#include <math.h>
#include <iostream>
int arrayer(int ar[101], int i);
int main()
{
srand(time(0));
int ar[101];
for (int i = 1; i < 101; ++i)
{
int arrayer(int ar[101], int i);
}
return 0;
}
int arrayer(int ar[101], int i) {
ar[i] = rand() % 100 + 1;
if (ar[i] < 10) {
std::cout << i << ": " << "0" << ar[i] << std::endl;
}
else {
std::cout << i << ": " << ar[i] << std::endl;
}
return ar[i];
}

Answer 1

You're calling and declaring the function incorrectly. This is what it should look like:
#include "stdafx.h"
#include <time.h>
#include <math.h>
#include <iostream>
int arrayer(int ar[101], int i);
int main() {
srand(time(0));
int ar[101];
for (int i = 1; i < 101; ++i)
{
arrayer(ar, i);
}
return 0;
}
int arrayer(int* ar, int i) {
ar[i] = rand() % 100 + 1;
if (ar[i] < 10) {
std::cout << i << ": " << "0" << ar[i] << std::endl;
}
else {
std::cout << i << ": " << ar[i] << std::endl;
}
return ar[i];
}
Notice also that you're not using the return value, so you can omit that if it's not going to ever be used.
EDIT: You can actually replace the if-else for printing the value with this:
std::cout << i << ": " << setw(2) << setfill('0') << ar[i] << std::endl;
You will need to include <iomanip> to do that.

Answer 2

You are passing the arguments in a very wrong manner, the way of calling a function is way more different than declaring or defining it. You need to call the function as :
arrayer(ar, i);
Just pass the address of the array, ar as it is and the variable i.
Also, its better to have the function arrayer return void rather than int, since the array gets modified when passed to the function and the value gets printed in the function itself, so you don't need to return anything.

Answer 3

Well there is nothing in output because you never call the arrayer function!
In your loop, you just declare again the function. As the new declaration is compatible with the previous one, the compiler accepts it without error.
In order to actually call the function, you need an function call expression:
for (int i = 1; i < 101; ++i)
{
arrayer(ar, i);
}