Erasure of Generic Information

• Type Erasure - Java does not keep record all versions of LinkedList<T> as separate types.

  • i.e. We cannot check like this:

    if (s instanceof LinkedList<String> ) { ... }
    

• At runtime, all type variables are promoted to Object.

  • LinkedList<T> becomes LinkedList<Object> .

• Or the Upper Bound, if one is available

  • LinkedList<? extends Shape> becomes LinkedList<Shape>

• So since no information of T is preserved, we cannot use T in expressions like: if (o instanceof T) {...}

What does type erasure mean for us??

• Type erasure now means that the code which implements a comparison below always returns True.

  o1 = new LinkedList<Employee>();
  o2 = new LinkedList<Date>();
  
  if(o1.getClass() == o2.getClass()){
  	//True -> this block is executed always.
  }
  

• Now as a consequence of this, the following overloading is illegal!!

  public class Example{
  	public void printlist(LinkedList<String> strList){ ... }
  	public void printlist(LinkedList<Date> datelist){ ... }
  }
  

  • Since now, after type erasure, both functions will have the same signature.

Wrapper Classes

• Now, during type erasure, all types variables are promoted to Object. (Look Here)

• So, the basic(primitive) types like int, float, … are not compatible with Object

  • Thus, we cannot use LinkedList<T> as LinkedList<int> , LinkedList<double>,…

• So Java provides Wrapper classes for each basic types:

  Basic Type	Wrapper Class
  byte	Byte
  short	Short
  int	Integer
  long	Long

  Basic Type	Wrapper Class
  float	Float
  double	Double
  boolean	Boolean
  char	Character

<aside> 💡 All wrapper classes other than Boolean and Character extend the class Number.

</aside>

• Converting from basic type to wrapper class and back:

  int x = 5;
  Integer myx = Integer(x);
  int y = myx.intValue();
  

• Autoboxing - implicit conversion between base types and their wrapper types.

  int x = 5;
  Integer myx = x;
  int y = myx;
  

<aside> 💡 Use Wrapper types in generic functions.

</aside>