Strange Generics Behavior

[Benjamin Lerman]

Hi all,

I encounter a strange behavior with generics that I do not understand.

I have the following interfaces:

public interface Int1<E> {
}

and

public interface Int2<E> extends Int1<Integer> {
}

and I have somewhere the following lines:

void foo(Int2 o) {
Int1<Integer> p = o;
}

The java compiler gives the following warning:

Type safety: The expression of type Int2 needs unchecked conversion to
conform to Int1<Integer>

I do not understand why this warning exists. Moreover it prevents the
compiler to understand that an Int2<E> is also an Int1<Integer>, so it
prevents a lot of type checking.

Is there any solution ?

Thanks.

Benjamin Lerman

 

[hiwa]

}

There can't be such.
It should be:

interface Int2 extends Int1<Integer> {

}

 

[Benjamin Lerman]

There can't be such.
It should be:

interface Int2 extends Int1<Integer> {

}

Why?

Note that E in Int2 is not Integer. I want to have something like:

public class O1 implements Int2<Float> {
}

Benjamin

 

[Chris Uppal]

Type safety: The expression of type Int2 needs unchecked conversion to
conform to Int1<Integer>

I do not understand why this warning exists. Moreover it prevents the
compiler to understand that an Int2<E> is also an Int1<Integer>, so it
prevents a lot of type checking.

I think the warning is spurious. I suspect that there's a bug somewhere.

The compiler (as far as I know) emits this warning when it is unable to prove
to itself that an operation is safe /and/ cannot emit a runtime checkcast
operation (because the types are identical after erasure). In this case there
is no checkcast (correctly), but I can't see why it shouldn't realise that
anything which conforms to Int2<anything> necessarily also conforms to

Is there any solution ?

Just ignore it.

-- chris

 

[Simon]

Hi,

public interface Int1<E> {
}

and

public interface Int2<E> extends Int1<Integer> {
}

and I have somewhere the following lines:

void foo(Int2 o) {
Int1<Integer> p = o;
}

The java compiler gives the following warning:

Type safety: The expression of type Int2 needs unchecked conversion to
conform to Int1<Integer>

I'm not sure why you get that warning, but you can avoid it if you declare foo
like this:

void foo(Int2<Object> o)

You can reproduce the same warning with the following simpler code that doesn't
use inheritance:

Int1 o = null;
Int1<Object> p = o;

I wonder what the difference is between Int1 and Int1<Object>. Is there any
example where this is significant or rather where using Int1 instead of
Int1<Object> causes any problems/exceptions?

 

[Piotr Kobzda]

Is there any solution ?

This:

void foo(Int2 o) {
Int1<Integer> p = (Int2<?>) o;
}

or better:

void foo(Int2<?> o) {
Int1<Integer> p = o;
}


Java compilers treat raw types specially.


Regards,
piotr

 

[Hendrik Maryns]

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Simon schreef:

Hi,

Indeed, for the same reason why you get this warning when you use List

I'm not sure why you get that warning, but you can avoid it if you declare foo
like this:

void foo(Int2<Object> o)

Well, I would declare foo like something more meaningful than

You can reproduce the same warning with the following simpler code that doesn't
use inheritance:

Int1 o = null;
Int1<Object> p = o;

I wonder what the difference is between Int1 and Int1<Object>. Is there any
example where this is significant or rather where using Int1 instead of
Int1<Object> causes any problems/exceptions?

There is a huge difference between Collection<Object> and just
Collection. For example Collection<Integer> is an heir of Collection,
but not of Collection<Object>. (Interestingly, in Eiffel this is not
the case. I wonder how they handle the problems that arise out of this.
They had generics from the start, so the whole type erasure stuff does
not exist.)

H.
- --
Hendrik Maryns

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