coding style - Can I get Scala to infer the Option type here? -

i want call generic function f[x](...), , in case x happens option[y]. try pass both some[...] , none function expects x, scala insists x of type some[y].

  def flattenoptionmap[a, b](input : map[a, option[b]]) : option[map[a, b]] = {     input.foldleft[option[map[a,b]]] (some(map.empty)) {       case (_, (_, none)) => none       case (none, (_, _)) => none       case (some(acc), (key, some(value))) => some(acc + (key -> value))     }   } 

in example, had explicitly specify option[map[a,b]] should used generic type foldleft. necessary type information contained in context, , typing cumbersome types option[map[a,b]] more necessary in opinion drastically reduces readability of code.

is there way scala infer type after all, or otherwise avoid copy-pasting whole type?

when use option(map.empty[a, b]) start value of foldleft, scala infer correct type wrote in comments (and beefyhalo in answer).

i add, if open using scalaz, can use sequence function.

import scalaz._ import scalaz._   val mapsome = map(1 -> some("a"), 2 -> some("b")) val mapnone = map(1 -> some("a"), 2 -> some("b"), 3 -> none)  mapsome.sequence flattenoptionmap(mapsome) // option[map[int,string]] = some(map(1 -> a, 2 -> b))  mapnone.sequence flattenoptionmap(mapnone) // option[map[int,string]] = none