Why Generic type can not instantiated?

Question

Here is my question.

class Gen<T> {
    T ob;
    Gen() {
        ob = new T(); // Illegal!!!
    }
}

Why is it illegal? Could you please explain it.

@DeadMG Are there languages which would allow some analogous code to compile and still guarantee no runtime errors in case when T happens to be something which cannot be instantiated with no-arg constructor (or at all)? — scriptin, Jun 4, 2016 at 18:14
@scriptin C# has new() constraint which allows to instantiate type parameters with no args constructors msdn.microsoft.com/en-us/library/sd2w2ew5.aspx — Esben Skov Pedersen, Jun 4, 2016 at 21:53

Andy Hunt · Accepted Answer · 2016-06-07 10:00:56Z

16

This is impossible because of the following 2 reasons.

There is no guarantee that T has a no-args constructor (and for that matter isn't an interface or abstract class)
Due to type erasure (required for backwards compatibility) the Type of T is known at compile time but not at run time, so what to construct wouldn't be known.

An answer may be to take a T factory in the constructor. Then Gen can request new Ts to its heart content.

Andy Hunt

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answered Jun 4, 2016 at 17:56

Richard Tingle

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1

After compile it will be replaced by Object . ob = new Object();
– alakhya
Jun 4, 2016 at 19:12
1

No it won't, you can't cast Object to T. It would break.
– Zalomon
Jun 9, 2016 at 14:18
"the Type of T is known at compile time" It is not know at compile time. When compiling this method, the compiler just sees a T, and checks to make sure that the code works for all possible T. It doesn't know what T is.
– user102008
Oct 21, 2016 at 21:53
@user102008 That cannot be the case. If so List<String> test=new ArrayList<>(); test.add(new Donkey()) would not be a compile error. Possibly we are meaning two different compile times, the compile time for the ArrayList itself and the compile time for the calling context
– Richard Tingle
Oct 22, 2016 at 10:48
@RichardTingle: But that's not in the code for List or ArrayList#add. That's in the code that uses it. When compiling ArrayList, the compiler doesn't know anything about what T is.
– user102008
Oct 22, 2016 at 16:42

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