forall in a lambda abstraction
20 May
2005
20 May
'05
7:26 p.m.
Hello, what's wrong with the code of test1 below. Ghci does complain with an error message that I don't understand. Thanks Jan id x = x test1 = \f ::(forall a. a -> a) -> (f True,f 'a') test = \f -> (f True,f False) h :: (forall a. a->a) -> (Bool,Char) --h :: (a.a -> a) -> (Bool,Char) h id = (id True, id 'a')
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Albert Lai -
Scott J.