Why can't Scala infer the type parameter in this example?

You will sometimes get better results if you use multiple parameter lists: def connectT(output: OutputT)(input: Input_ >: T) = { output. Input = Some(input) input. Output = Some(output) } connect(out)(in) and indeed in this case, it works.

You will sometimes get better results if you use multiple parameter lists: def connectT(output: OutputT)(input: Input_ >: T) = { output. Input = Some(input) input. Output = Some(output) } connect(out)(in) ...and indeed in this case, it works.

"sometimes get better results" doesn't sound very deterministic! – oxbow_lakes Jul 27 '09 at 12:03 2 Sadly, the type inferencer isn't spec'ed, and sometimes isn't deterministic either. – Jorge Ortiz Jul 28 '09 at 3:17.

I may be totally wrong, but I think the problem is when you tie together the input and the output: the input has an output restricted to a subtype of T, but the output has an input restricted to a supertype of T, the only type that can satisfy both conditions is T. InputT -> Output _ Input _ >: Q When you create the input with the output (replacing Q with _ Input _ >: _ Input.

In my example, String and AnyRef satisfy my constraints. The Output produces Strings and requires an Input that consumes some supertype of String. The Input consumes AnyRefs and requires an Output that produces some subtype of AnyRef.

Since String is a subtype of AnyRef, the constraints are satisfied. – Jay Conrod Jul 27 '09 at 18:26 The problem is that there is exactly one appropriate type parameter for connect, and that is the type parameter of the Output. When I specify this explicitly, it works fine.

If I don't, it tries to use the type parameter of the Input, and I get a type error on the Output argument. – Jay Conrod Jul 27 '09 at 18:28 Sorry, I misread the Input_ >: T in the signature of connectT( – GClaramunt Jul 28 '09 at 16:30.

Actually scala type inferene can't handle "recursion" for now. So if type of argument can be infered only in collocation with other argument scala fail to inference. But if you use differnt argument list scala f(a)(b)(c,d)will infer types list by list, so it usally works better.PS It is oversimplified, but can get you some clue.

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