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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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