Instaciate a generic type

Asked Sep 28 '13 at 17:55

Active Sep 28 '13 at 17:59

Viewed 53 times

Consider the following generic class:

class MyClass<T>{
    T field;//ok
    public void method (T t){...}//ok
    public <E> MyClass(E[] e){
        E ee= new E();// compile error
        T t= new T();// compile error
    }
}

Why we cant instanciate of T and E? We know all about of types T,E at run time. I think that the type parameters T,E can be considered as a some reference type.

asked Sep 28 '13 at 17:55

St.Antario

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`T` and `E` don't exist at runtime. – Sotirios Delimanolis Sep 28 '13 at 17:57
@SotiriosDelimanolis that should be an answer :P – Zavior Sep 28 '13 at 17:58
1

Read about `type erasure` in Generics. – anubhava Sep 28 '13 at 17:58
@Sotirios Delimanolis After type erasure at compile time we have that `E,T` are defined to `Object`. What problems to instanciate of `Object`? I dont undestand. – St.Antario Sep 28 '13 at 18:04
`T` isn't necessarily going to be `Object`. – Sotirios Delimanolis Sep 28 '13 at 18:05
@Sotirios Delimanolis Why not necessarily? `T,E` without bounds, hence after type erasure `T,E` are defined to `Object` at compile time – St.Antario Sep 28 '13 at 18:08
That's not how it works. And even if it did, it would immediately fail at runtime. See the linked question in a previous comment. – Sotirios Delimanolis Sep 28 '13 at 18:11
@Sotirios Delimanolis Does at compile time `new` operator requires a specific type? Compiler checked this before type erasure, but at the type checked. – St.Antario Sep 28 '13 at 18:30

Instaciate a generic type

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