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by jandrese 3376 days ago
No, it's a mismatch between what the compiler accepts and what the JVM can execute.

The interesting part to me is that it seems that Java would be perfectly capable of differentiating between methods by return type if the compiler was tweaked slightly. Is there a reason why this isn't a formal language feature?
1 comments
barahilia 3376 days ago
As explained in the post, function call is a full expression for which appropriate function should be found. If there 2 functions with the same name and param types it would be impossible to compile such an expression in its own.
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masklinn 3376 days ago
Not necessarily, you'd just need a syntactic mechanism to disambiguate.
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samatman 3376 days ago
And have one, since Java either throws away the return value, or assigns it to a variable the type of which is known.

Edit: here's the edge case though. You can call a function and use the return value directly as a parameter: foo(bar()). It's possible to have two foo that take both possible bar return types, at which point the compiler is stuck.

It could require a cast in this instance, however. The more I think about this the more I wonder why this isn't possible.

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masklinn 3375 days ago
> here's the edge case though.

That's no more an edge case than the cases where you throw away the return value or you're binding to an ambiguous type e.g. `A getFoo()`, `B getFoo()`, `Object foo = getFoo()`.

> The more I think about this the more I wonder why this isn't possible.

The Java spec does not say, for C++ Stroustrup states it's

> to keep resolution for an individual operator or function call context-independent.

the Java reason is likely also some sort of Principle of Least Surprise claim.

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