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I was going through TypeErasure topic at http://download.oracle.com/javase/tutorial/java/generics/erasure.html which says that compiler removes all information related to type parameters and type arguments within a class or method.

Now considering the code below

public class Box<T> {
    private T t; // lineA, T stands for "Type" 

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

    public T get() { // lineD
        return t; // lineE
    }
}

Now inside main method I have below code snippet

Box<String> box1 = new Box<String>(); // line1
box1.add("Scott"); // line2
String s1tr=box1.get(); // line3

Box<Integer> box2 = new Box<Integer>(); // line4
box2.add(1); // line5
Integer k=box2.get(); // line6

Now in above code (in Box class and main method) what are the changes compiler will make and at which line?

As the link says that compiler removes all information related to type parameters and type arguments within a class or method, when compiler will compile Box class, will it remove all T,<String>,<Integer> occurences from Box class and main method respectively? If yes, what will be the compiled code after Removing T?

Guido
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M Sach
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  • Line 4 in your example won't work. It will trigger a compiler error because the types (`Box` and `Box`) do not match. – joschi Sep 10 '11 at 11:43
  • For the most part Type Erasure means it makes the byte code not need to know the generic type (some extra code is generated) however in many ways it like calling Comment Erasure a feature. ;) – Peter Lawrey Sep 10 '11 at 12:14
  • Type erasure isn't the feature, "generics" is. Type erasure is a design choice made when implementing generics, trading expressiveness for binary compatibility. – millimoose Sep 10 '11 at 12:54

2 Answers2

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Type erasure happens at compile time. Java compiler removes those generic type information from source and adds casts as needed and delivers the byte code. Therefore the generated byte code will not have any information about type parameters and arguments. It will look like a old java code without generics. There is no way to determine the value of T at runtime, because that information is removed before the code is compiled.

Anthony
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  • For the sake of completeness, the code will have *some* information that's left over from type bounds, and "super type tokens" (http://gafter.blogspot.com/2006/12/super-type-tokens.html). Meaning that in some rather specific cases you can recover the value of T. – millimoose Sep 10 '11 at 11:58
  • Right, I guess what I meant to say was there is no 'reliable' way to determine the types. – Anthony Sep 10 '11 at 12:01
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In the absence of type bounds like in your example code, what will (roughly) happen that in the Box class, all instances of all references to T in method code will be replaced by Object. In the code using the Box class, all instances of <String>, <Integer> and such will be removed.

An exhaustive in-depth article on Java Generics is Angelika Langer's FAQ - it also explains the effect of type bounds and a lot more.

millimoose
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  • Yes agreed that all instances of will be removed, and all references to T will be replaced by Object.But still compiler will have some meta data information after compiling the Box class so that when it is refered by other class like main method in above example it apply compile time checks .Right? – M Sach Sep 10 '11 at 12:22
  • @Mohit: You're probably correct, type erasure affects method code not signatures. I'll edit my answer to correct this. – millimoose Sep 10 '11 at 12:44