[Haskell-cafe] forall in a lambda abstraction

Albert Lai trebla at vex.net
Fri May 20 13:34:00 EDT 2005


"Scott J." <jscott at skynet.be> writes:

> what's wrong with the code of test1 below. Ghci does complain with
> an error message that I don't understand.

> test1 = \f ::(forall a. a -> a) -> (f True,f 'a')  

test1 = \(f :: forall a. a -> a) -> (f True,f 'a')



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