Doug Gregor ad2fa39eaa Treat deducible generic parameter types as unresolved types, because ...

they need to be substituted away for any type containing them to be
useful. A nice side-effect of this is that we'll actually properly
walk into the source type when a generic function returns, e.g., a
tuple type that involves type parameters not resolved by the input
arguments.


Swift SVN r2302

2012-07-03 23:23:25 +00:00

..

swift

Treat deducible generic parameter types as unresolved types, because

2012-07-03 23:23:25 +00:00