[Haskell] Re: Polymorphic types without type constructors?

2 Feb 2005


      Hi,

Shin-Cheng Mu wrote:
...
exp2 = \m::N -> \n::N -> \f::(b -> b) -> \b::b ->
           n [b->b] (m[b]) f b
Oops.. it should be

   exp2 = \m::N -> \n::N -> \f::(b -> b) -> \b::b ->
             n [b] (m[b]) f b

sincerely,
Shin-Cheng Mu