java extending Generic types ambiguous method call

Question

I have run into a problem which is hard for me to understand why this happens.

I have a generic class B which extends generic class A and i have two methods accepting A and B. From method accepting A i want to call method with paramter B but this results in compilation error:

Ambiguous method call. Both method (A) in Main and method (B) in Main match

Here is a code snippet:

public class Main {

private static class A<T> {

}

private static class B<T> extends A {

}

public String method(A<String> arg) {
    return method((B<String>) arg));
}

public String method(B<String> arg) {
    return "some result";
}
}

However if I remove generic type from type A it compiles:

public class Main {

private static class A<T> {

}

private static class B<T> extends A {

}

public String method(A arg) {
    return method((B<String>) arg));
}

public String method(B<String> arg) {
    return "some result";
}
}

What is the reason of that ?

Edit: This issue is somehow related to java 8 because this code is compilable with JDK7

Possible duplicate of Compiler error : reference to call ambiguous OR this — Naman
– Naman, Commented Sep 21, 2017 at 9:57
Well, it does not quite make sense on what you are doing. Anyway: A<String> is-NOT-a A<Object> . As you declare your B<T> extends A, which means B<T> extends A<Object>, which means, B<WhateverType> is-a A<Object>. Given A<Object> is-NOT-a A<String>, B<String> is-NOT-a A<String>, so the compiler is actually complaining it correctly by avoiding you to cast from A<String> to B<String> as it will never be lega — Adrian Shum
– Adrian Shum, Commented Sep 25, 2017 at 2:53

Alberto S. · Accepted Answer · 2017-09-21 09:52:45Z

2

Your class B extends A but it's not specifying its bounds nor any generic info

What I mean is that your line

private static class B<T> extends A { ... }

is equivalent to

private static class B<T> extends A<Object> { ... }

By changing it to

private static class B<T> extends A<T> { ... }

Your code will compile

answered Sep 21, 2017 at 9:52

Alberto S.

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1 Comment

homik Over a year ago

You are right it will compile but what if i don't want them to share the types ? I cast the parameter to B to explicitly call the other method but it looks like compiler ignores it

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