> > Turns out that Rémi's example exposes the difference between the wildcard > approach and the type-parameter approach. Returning to the example, > > Optional<Integer> oi = Optional.empty(); > Function<Number, Optional<StringBuilder>> fm = n -> Optional.empty(); > Optional<CharSequence> ocs = oi.flatMap(fm); > > If the flatMapper function itself has a wildcard type, for example, > > Function<Number, Optional<? extends CharSequence>> fm = n -> > Optional.empty(); > > then this will still work with the wildcard approach but fail with the > type-parameter approach. As Rémi also pointed out, a wildcarded type can > result from the capture of a type with a wildcarded type parameter. > > Based on this, I believe the nested wildcard approach to be the correct one. > > s'marks >
Yes, this argument is unchallengeable. I'm convinced now - though not happy. Kind regards, Stefan
