When does the Scala compiler really need the type information of parameters of anonymous functions?
For instance, given this function:
def callOn[T,R](ta
The trick to type inference is to consider it as an process of iterative refinement. Each parameter block can be used to infer some of the type parameters, which can then be used in subsequent blocks. So take the following definition:
def chain[T,A,B](x: T)(fn1: T=>A)(fn2: A=>B) = fn2(fn1(x))
called as:
chain(2)(_*10)("xxx"+_)
So how is this inferred? First, we start with the block (2)
which is known to have the type Int
. Substituting that back into the T
parameter we get:
def chain[A,B](x: Int)(fn1: Int=>A)(fn2: A=>B) = fn2(fn1(x))
The next parameter block is (_*10)
, where we now know the type of the placeholder _
to be Int
... and multiplying an Int
by an Int
gives another Int
. This is true of the returned type even if an overflow occurrs; at the extreme end it may throw an exception, but exceptions have the type Nothing
which is a subclass of everything else in the type system, so we can still say Nothing
is-an Int
and the inferred type of Int
is still valid.
With A
inferred, the method becomes:
def chain[B](x: Int)(fn1: Int=>Int)(fn2: Int=>B) = fn2(fn1(x))
Leaving only B
which can be inferred from ("xxx"+_)
. As String + Int
is a String
, the method is now:
def chain(x: Int)(fn1: Int=>Int)(fn2: Int=>String) = fn2(fn1(x))
As the return type of the method comes direct from fn2
, that can also be explicitly shown for completeness:
def chain(x: Int)(fn1: Int=>Int)(fn2: Int=>String): String = fn2(fn1(x))
There you have it, all types safely resolved, and the method proven to be statically valid.
In your case, you need the type T
to be inferred before it's possible to infer R
from the type T=>R
. To do this you must split out the parameters into two distinct blocks, writing the method in a curried form:
def callOn[T,R](target: T)(f: (T => R)) = f(target)