Java Generics - Generic Methods Erasure


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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("Integer Value :" + box.get());
   }
   
   private static <T> void printBox1(T box) {
      System.out.println("String 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("Integer Value :" + box.get());
   }
	
   //Unbounded Types Erasure
   private static void printBox1(Object box) {
      System.out.println("String 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

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