binary serach for infinite

// Java program to demonstrate working of  
// an algorithm that finds an element in an  
// array of infinite size 
  
class Test 
{ 
    // Simple binary search algorithm 
    static int binarySearch(int arr[], int l, int r, int x) 
    { 
        if (r>=l) 
        { 
            int mid = l + (r - l)/2; 
            if (arr[mid] == x) 
                return mid; 
            if (arr[mid] > x) 
                return binarySearch(arr, l, mid-1, x); 
            return binarySearch(arr, mid+1, r, x); 
        } 
        return -1; 
    } 
      
    // Method takes an infinite size array and a key to be 
    // searched and returns its position if found else -1. 
    // We don't know size of arr[] and we can assume size to be 
    // infinite in this function. 
    // NOTE THAT THIS FUNCTION ASSUMES arr[] TO BE OF INFINITE SIZE 
    // THEREFORE, THERE IS NO INDEX OUT OF BOUND CHECKING 
    static int findPos(int arr[],int key)  
    { 
        int l = 0, h = 1; 
        int val = arr[0]; 
  
        // Find h to do binary search 
        while (val < key) 
        { 
            l = h;     // store previous high 
            //check that 2*h doesn't exceeds array  
            //length to prevent ArrayOutOfBoundException 
            if(2*h < arr.length-1) 
               h = 2*h;              
            else
               h = arr.length-1; 
              
            val = arr[h]; // update new val 
        } 
  
        // at this point we have updated low 
        // and high indices, thus use binary  
        // search between them 
        return binarySearch(arr, l, h, key); 
    } 
  
    // Driver method to test the above function 
    public static void main(String[] args)  
    { 
        int arr[] = new int[]{3, 5, 7, 9, 10, 90,  
                            100, 130, 140, 160, 170}; 
        int ans = findPos(arr,10); 
          
        if (ans==-1) 
            System.out.println("Element not found"); 
        else
            System.out.println("Element found at index " + ans); 
    } 
}