Wash in detail?

Question

I am learning generics in Java. I learned that during compilation, erasing erases all the general information.

Here I learned that all T is replaced by Object.

Integer v = display(2); String d = display("3"); public <T> T display(T i){ return i; } 

I hope he turns into

 Integer v = (Integer)display(2); String d = (String) display("3"); public Object display(Object i){ return i; } 

So, is my assumption true?

Thanks.

Answer 1

You basically have this. The compiler uses common parameter types to verify the code, but the generated bytecode replaces another class.

For an unrelated common parameter such as <T> , this will be Object . However, if the declaration was <T extends Comparable> , the replaced class would be Comparable instead of Object .

Edit : Excellent Java Generics information, albeit very dense with piecemeal information:

Java Generics Frequently Asked Questions - Frequently Asked Questions
http://www.angelikalanger.com/GenericsFAQ/JavaGenericsFAQ.html

Answer 2

information about the generic type is not stored in the compiled class; it is used only at compile time to check the type's security.

 Integer v = display(2); 

The Java compiler knows that this method is “general”, and it returns the same type as its argument, in this case it is Integer, and the result is assigned to Integer, so everything is fine.

 String d = display("3"); 

completely the same trick. But in this case:

 Long v = display(2); 

Java will tell you that it cannot use, because 2 must be Integer, and Java cannot give it Long, so you need to write:

 Long v = display(2L); 

As I said, no information is stored in the compiled class, so potentially at runtime you can enter a List<Integer> value of type String , and no exception will be thrown, but when you read this String and try to assign a variable of type Integer , you get it ... Well, pretty hard;)

Answer 3

I guess, yes. Generics in Java is a good compiler.