Java Generics - Generic Methods Erasure

Java Compiler replaces type parameters in generic type with Object if unbounded type parameters are used, and with type if bound parameters are used as method parameters.

Example

package com.tutorialspoint;

public class GenericsTester {
   public static void main(String[] args) {
      Box<Integer> integerBox = new Box<Integer>();
      Box<String> stringBox = new Box<String>();

      integerBox.add(new Integer(10));
      stringBox.add(new String("Hello World"));
      
      printBox(integerBox);
      printBox1(stringBox);
   }
   
   private static <T extends Box> void printBox(T box){
      System.out.println("Value: " + box.get());
   }
   
   private static <T> void printBox1(T box){
      System.out.println("Value: " + ((Box)box).get());
   }
}

class Box<T> {
   private T t;

   public void add(T t) {
      this.t = t;
   }

   public T get() {
      return t;
   }   
}

In this case, java compiler will replace T with Object class and after type erasure,compiler will generate bytecode for the following code.

package com.tutorialspoint;

public class GenericsTester {
   public static void main(String[] args) {
      Box integerBox = new Box();
      Box stringBox = new Box();

      integerBox.add(new Integer(10));
      stringBox.add(new String("Hello World"));
      
      printBox(integerBox);
      printBox1(stringBox);
   }
	
   //Bounded Types Erasure
   private static void printBox(Box box){
      System.out.println("Value: " + box.get());
   }
	
   //Unbounded Types Erasure
   private static void printBox1(Object box){
      System.out.println("Value: " + ((Box)box).get());
   }
}

class Box {
   private Object t;

   public void add(Object t) {
      this.t = t;
   }

   public Object get() {
      return t;
   }   
}

In both case, result is same −

Output

Integer Value :10
String Value :Hello World