Get size of an array in C++ 17 using pass by reference - c++17

Question

Is there a way to find the size of an array, using pass by reference. The size function works well inside the main function.
#include <iostream>
#include <iterator>
using namespace std;
int hello(int arr[]){
cout<<arr<<endl;
//cout<<size(arr)<<endl;
}
int main(){
int arr[] = {1,2,3,4,6,7,2};
hello(arr);
cout<<arr<<endl;
cout<<size(arr)<<endl;
}

Answer 1

You are passing array by value so hello function takes as its parameter decayed type - int*.
When you pass array by reference, you need to pass one additional non-type template parameter which is length of array and is deduced, therefore inside hello you don't even use std::size to get array's length, just use N:
template<class T, size_t N>
int hello(T (&arr)[N])
{
cout<< size(arr) <<endl; // 7
cout << N << endl; // 7
return 0;
}

Answer 2

Once you pass an array to a function, it becomes a pointer, and there is no way to get the size from that (it will always report 4 or 8 bytes - the size of a pointer value!).
If you need to process an array in a func, you will need to pass in the size as an argument to the function.
#include <iostream>
#include <iterator>
using namespace std;
int hello(int arr[], size_t count)
{
for(size_t i = 0; i < count; ++i)
{
std::cout << arr[i] << std::endl;
}
}
int main(){
int arr[] = {1,2,3,4,6,7,2};
hello(arr, sizeof(arr) / sizeof(int));
return 0;
}
The better alternative in C++, is to use std::vector instead. (which will store the size as a member variable)
#include <iostream>
#include <iterator>
#include <vector>
using namespace std;
void hello(const std::vector<int>& arr)
{
for(size_t i = 0; i < arr.size(); ++i)
{
std::cout << arr[i] << std::endl;
}
}
int main(){
std::vector<int> arr = {1,2,3,4,6,7,2};
hello(arr);
return 0;
}