We can find size of an array using sizeof operator as shown below.

// Finds size of arr[] and stores in 'size'
int size = sizeof(arr)/sizeof(arr[0]);

Can we do the same without using sizeof operator?

Method 1 (Writing our own sizeof)
Given an array (you dont know the type of elements in the array), find the total number of elements in the array without using sizeof operator?

One solution is to write our own sizeof operator (See this for details)

// C++ program to find size of an array by writing our
// sizeof
#include <bits/stdc++.h>
using namespace std;

// User defined sizeof macro
# define my_sizeof(type) ((char *)(&type+1)-(char*)(&type))

int main()
{
	int arr[] = {1, 2, 3, 4, 5, 6};
	int size = my_sizeof(arr)/my_sizeof(arr[0]);

	cout << "Number of elements in arr[] is "
		<< size;

	return 0;
}

Output :

Number of elements in arr[] is 6

Method 2(Using a pointer hack)
The following solution is very short when compared to the above solution. Number of elements in an array A can be found out using the expression

int size = *(&arr + 1) - arr;

// C++ program to find size of an array by using a
// pointer hack.
#include <bits/stdc++.h>
using namespace std;

int main()
{
	int arr[] = {1, 2, 3, 4, 5, 6};
	int size = *(&arr + 1) - arr;
	cout << "Number of elements in arr[] is "
		<< size;
	return 0;
}

Output :

Number of elements in arr[] is 6

How does this work?
Here the pointer arithmetic does its part. We don’t need to explicitly convert each of the locations to character pointers.

&arr ==> Pointer to an array of 6 elements.
         [See this for difference between &arr
          and arr]   

(&arr + 1) ==> Address of 6 integers ahead as
               pointer type is pointer to array
               of 6 integers.

*(&arr + 1) ==> Same address as (&arr + 1), but 
                type of pointer is "int *".

*(&arr + 1) - arr ==> Since *(&arr + 1) points 
                   to the address 6 integers
                   ahead of arr, the difference
                   between two is 6.