[Haskell] Re: Polymorphic types without type constructors?

Shin-Cheng Mu scm at ipl.t.u-tokyo.ac.jp
Tue Feb 1 22:25:01 EST 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



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