dichotomic search c 2b 2b

1#include<iostream> 
2using namespace std; 
3int binarySearch(int arr[], int p, int r, int num) { 
4   if (p <= r) { 
5      int mid = (p + r)/2; 
6      if (arr[mid] == num)   
7         return mid ; 
8      if (arr[mid] > num)  
9         return binarySearch(arr, p, mid-1, num);            
10      if (arr[mid] < num)
11         return binarySearch(arr, mid+1, r, num); 
12   } 
13   return -1; 
14} 
15int main(void) { 
16   int arr[] = {1, 3, 7, 15, 18, 20, 25, 33, 36, 40}; 
17   int n = sizeof(arr)/ sizeof(arr[0]); 
18   int num = 33; 
19   int index = binarySearch (arr, 0, n-1, num); 
20   if(index == -1)
21      cout<< num <<" is not present in the array";
22   else
23      cout<< num <<" is present at index "<< index <<" in the array"; 
24   return 0; 
25}

1using namespace std; 
2  
3// A recursive binary search function. It returns 
4// location of x in given array arr[l..r] is present, 
5// otherwise -1 
6int binarySearch(int arr[], int l, int r, int x) 
7{ 
8    if (r >= l) { 
9        int mid = l + (r - l) / 2; 
10  
11        // If the element is present at the middle 
12        // itself 
13        if (arr[mid] == x) 
14            return mid; 
15  
16        // If element is smaller than mid, then 
17        // it can only be present in left subarray 
18        if (arr[mid] > x) 
19            return binarySearch(arr, l, mid - 1, x); 
20  
21        // Else the element can only be present 
22        // in right subarray 
23        return binarySearch(arr, mid + 1, r, x); 
24    } 
25  
26    // We reach here when element is not 
27    // present in array 
28    return -1; 
29} 
30  
31int main(void) 
32{ 
33    int arr[] = { 2, 3, 4, 10, 40 }; 
34    int x = 10; 
35    int n = sizeof(arr) / sizeof(arr[0]); 
36    int result = binarySearch(arr, 0, n - 1, x); 
37    (result == -1) ? cout << "Element is not present in array"
38                   : cout << "Element is present at index " << result; 
39    return 0; 
40}