Generic Binary Search - JAVA -

Question

I dont know if the functionality of my code is the correct one for binary search BUT this is not my question, I want to solve it myself.

My problem is in testing its functionality while I get the following errors:

I really dont know how to solve this issue. Please help me!

My code:

public class BinarySearchGeneric<T extends Comparable<T>>{

public int search(T[] array, T element){
    int start = 0;
    int end = array.length - 1;
    int mid = (start + end) / 2;

    while(!element.equals(array[mid]) && end != 0) {
        if (element.equals(array[mid])) {
            return mid;
        } else if (array[mid].compareTo(element) < 0) {
            end = mid - 1;
            mid = (start + end) / 2;
        } else if(array[mid].compareTo(element) > 0){
            start = mid + 1;
            mid = (start + end) / 2;
        }
    }
    return -1;
}
}

Main Method:

public class Main {
public static void main(String[] args) {

    int[] a = {1,2,3,4,5,6,7,8,9,10};

    BinarySearchGeneric binarySearchGeneric = new BinarySearchGeneric(a);
    System.out.println("BinarySearch Generic: " + binarySearchGeneric.search(a, 8));

    }
}

Answer 1

There are two compilation issues here:

1. There is no constructor of BinarySearchGeneric which takes a parameter, but you're trying to pass the parameter. Remove it:

   BinarySearchGeneric binarySearchGeneric = new BinarySearchGeneric();
   

2. int[] is not an acceptable parameter to a generic method expecting an array, because int is a primitive type, not a reference type, and so can't be used in generics. The solution is simply to declare an array of Integer, rather than int:

   Integer[] a = {1,2,3,4,5,6,7,8,9,10};
   

   The compiler converts these int literals to Integer instances automatically.

---

But there are more issues.

• You're declaring a variable of raw type. This basically switches off the compiler's type checking associated with that variable, making it likely that you will make a type error. Add the generic parameters:

  BinarySearchGeneric<Integer> binarySearchGeneric = new BinarySearchGeneric<>();
  

• Arrays and generics don't really play well together. Things would start get a bit messy if you declared a generic, comparable class:

  class GenericComparable<T> extends Comparable<T> { ... }
  

  and then tried to declare an array of GenericComparables to pass to binarySearchGeneric, since you can't directly create a generic array.

  It's much easier simply to avoid arrays, and use a List<T> instead:

  public int search(List<T> array, T element){
  

• Potentially, you have inconsistent behaviour, because you are mixing equals and compareTo in the search. Whilst compareTo should be consistent with equals (in the sense that a.compareTo(b) <=> a.equals(b), it isn't necessarily true.

  You can make the behaviour consistent by only using compareTo:

  int c = array[mid].compareTo(element);
  if (c == 0) {
    // ...
  } else if (c < 0) {
    // ...
  } else {
    // ...
  }