How to specify bound for a generic type with generic type in Scala?

Question

This question is about working with generic types and setting the bound, so please do not freak out because of the library I use. Treat it as an example.

I work with Scalala and use such types DenseMatrix[V], DenseVectorCol[V] and DenseVectorRow[V]. The common super type for them is Tensor[K,V]. Note that the Tensor has additional parameter K -- all the mentioned classes Dense... set this type K by themselves.

So I would like to write a method with argument which upper type bound is Tensor. I wrote such code for my method:

def validate[K,T <: Tensor[K,Double]](tensor : T) : T = ...

with such intention -- T has to be subtype of Tensor, and I know I work with Doubles all the time, so let it be Double, and for first type argument (K) get it from passed argument.

It does not work as I expected because I get error:

inferred type arguments [Nothing,DenseVectorCol[Double]] do not conform to method validate's type parameter bounds [K,T <: Tensor[K,Double]]

QUESTION: so how to extract this type K from the passed argument?

Note that Tensor gives you access to its type parameter K through its type member Domain, which might make it a little easier to get by with def validate[T <: X[_, Double]] ... (which will be inferred correctly). — Travis Brown
@TravisBrown, bingo -- wildcard _ does the job (I put Tensor instead of X of course). Why didn't you post a regular answer? Please do. — greenoldman

Travis Brown Travis Brown · Accepted Answer · 2012-09-15T23:47:50

If you don't care about K at all, you can use a wildcard:

def validate[T <: Tensor[_, Double]]: T = ...

Note that in some cases this wouldn't work (e.g., if you needed to return a K or otherwise use it in the method), but assuming this isn't one of those cases, this is a perfectly valid solution and the type inference will work out just fine.

How to specify bound for a generic type with generic type in Scala?

1 Answers