Why Java is not complaining about an ambiguous call?

Question

A service interface declares two methods which apparently do the same processing :

interface Service<T> {
    <R> R process(Function<? super T, ? extends R> function);
    T process(UnaryOperator<T> operator);
}

The service above is being called like below :

void process(Service<CharSequence> service) {
    service.process(sequence -> sequence.subSequence(0, 1));
}

Which one of the service methods are going to be called and why the compiler does not complain about an ambiguous call in this context?

T.J. Crowder · Accepted Answer · 2019-01-19T13:28:47.370

11

Method resolution chooses the most specific matching method when there are multiple possible matches. Since UnaryOperator<T> extends Function<T,T>, if that lambda matches it (and it does), it's more specific than Function<T, T> so the UnaryOperator overload will be used.

edited Jan 19 '19 at 13:28

answered Jan 19 '19 at 13:26

T.J. Crowder

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1

I totally forgot about the inheritance between a `UnaryOperator` and `Function`, thank you so much. Definitely, clears the ambiguity. – HPH Jan 19 '19 at 13:28
@HPH - To be fair, I had to look it up. :-) – T.J. Crowder Jan 19 '19 at 13:29

Why Java is not complaining about an ambiguous call?

1 Answers1