CODEWITHSUNDEEP
javaooppolymorphism
Mike Cooper
Mike Cooper

Reputation: 3048

Ambiguous Polymorphism?

In situations where two interfaces apply to an object, and there are two overloaded methods that differ only by distinguishing between those interfaces, which method gets called?

In code.

interface Foo {}
interface Bar {}

class Jaz implements Foo, Bar {}

void DoSomething(Foo theObject)
{
    System.out.println("Foo");
}

void DoSomething(Bar theObject)
{
    System.out.println("Bar");
}

Jaz j = new Jaz();
DoSomething(j);

Which method will get called? DoSomething(Foo) or DoSomething(Bar)? Neither is more specific than the other, and I see no reason why one should be called instead of the other, except that one is specified first/last.

EDIT: And with this type of code is it possible to force one or the other method?

Upvotes: 4

Views: 748

Answers (5)

Thorbjørn Ravn Andersen
Thorbjørn Ravn Andersen

Reputation: 75376

If the compiler is in doubt, it requires the programmer to cast the argument so it is not ambiguous anymore.

Try passing "null" as a parameter.

Upvotes: 0

Tommaso Taruffi
Tommaso Taruffi

Reputation: 9252

Is not ambiguos.

The signature of the method is: NAME and PARAMETERS

those methods are different.

void DoSomething(Foo theObject); //SIGNATURE IS: DoSomething,Foo

void DoSomething(Bar theObject); //SIGNATURE IS: DoSomething,Bar

Upvotes: 0

lc.
lc.

Reputation: 116468

I'm pretty sure the above won't compile. DoSomething(j) is an ambiguous call and will result in an error.

To get it to compile, you'd have to specifically cast j as a Foo or Bar when you call DoSomething, for example DoSomething((Foo)j).

Upvotes: 10

clemahieu
clemahieu

Reputation: 1429

That's an ambiguity in languages without renaming, most languages. Your best bet is to build two classes that implement the interface separately and make calls on the master object depending on which implementation you need to execute.

If you want to see a language that has function renaming check out Eiffel. http://dev.eiffel.com

Upvotes: 0

Clint
Clint

Reputation: 9058

This should be a compiler error.

This works:

DoSomething((Foo)j);
DoSomething((Bar)j);

Upvotes: 12

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