Re: [Haskell-cafe] why different type variables enforce different types?

Haskell does something implicit that you can't see. f :: a -> b really means f :: forall a. forall b. a -> b I think this is called universal quantification. It's like a c++ template. template<typename a, typename b> b function(a x) { return x; } So there needs to be a function f forall a and b. I can choose a=Int, b=Char which is impossible with f a = a Cheers Silvio