Compute average of two numbers without overflow

Given two numbers, a and b. Compute the average of the two numbers.

The well know formula (a + b) / 2 may fail at the following case :
If, a = b = (2^31) – 1; i.e. INT_MAX.
Now, (a+b) will cause overflow and hence formula (a + b) / 2 wont work

Improved Formula that does not cause overflow :

Average = (a / 2) + (b / 2) + (((a % 2) + (b % 2)) / 2)


Below is the implementation :

// C code to compute average of two numbers 
#include <stdio.h>
#define INT_MAX 2147483647
// Function to compute average of two numbers 
int compute_average(int a, int b) 
{ 
    return (a / 2) + (b / 2) + ((a % 2 + b % 2) / 2); 
} 
  
  int main() 
{ 
    // Assigning maximum integer value 
    int a = INT_MAX, b = INT_MAX; 
  
    // Average of two equal numbers is the same number 
    printf("Actual average : %d\n",INT_MAX); 
  
    // Function to get the average of 2 numbers 
    printf("Computed average : %d",compute_average(a, b)); 
  
    return 0; 
} 

Output:
Actual average: 2147483647
Computed average: 2147483647