Finding the number of occurrences of an integer in a sorted array using binary search

Extremely basic binary search implementation - C++

//
//  main.cpp
//  interview_questions
//
//  Created by Jonathan Nwosu on 25/03/2017.
//  Copyright © 2017 Jonathan Nwosu. All rights reserved.
//

#include <stdio.h>
#include <iostream>
using namespace std;

int binaryOccurance(int array[], int size, int x, bool searchFirst){

    int low, high, result, mid;
    
    low = 0;
    high = size - 1;
    
    result = -1;
    
    while (low <= high){
        
        mid = (low+high)/2;
    
        if(array[mid] == x){
    
        result = mid;
    
        if(searchFirst)
            high = mid -1; //Go on searching towards the left (lower indices)
         else
            low = mid +1; // Go on searching towards the right (higher indices)
            
        }
        else if(x < array[mid]) high = mid -1;
    
        else low = mid+1 ;
        
        }
    
return result;
    
}

int main() {

int array[] = {1,1,3,3,5,5,5,5,5,9,9,11};
    int size = sizeof(array)/sizeof(array[0]);
    int x;
    cout << "Number to find in array: " << endl;
    cin >> x;
    
    //cout << "Appears: ";
    
    //cout << findCount(array, size, x) << "times."<< endl;
    
   int firstIndex = binaryOccurance(array, size, x, true);
    
    if(firstIndex == -1){
        
        printf( "Count of %d is %d",x,0);
        
    } else {
        
        int lastIndex = binaryOccurance(array, size, x, false);
        printf("Count of %d is %d",x,lastIndex - firstIndex + 1);
        
    }

}