Benjamin Franksen wrote:
main = let ?b = True in use_b
--use_b :: (?g::Bool) => IO () use_b = print ?b
It isn't: ghc -fimplicit-params says
Unbound implicit parameter (?b :: a) arising from use of implicit parameter `?b' at TestBug.hs:4 In the first argument of `print', namely `?b' In the definition of `use_b': use_b = print ?b
It works if I uncomment the signature.
Using ghc-6.2.2, btw.
My question: Is this as it should be or is it a bug?
The Monomorphism Restriction. It's intended to prevent people from shooting themselves in the foot by defining things that look like constants (like use_b) but really are polymorphic values that still depend on some context and therefore are not constant. So in this case (use_b defined with no arguments), you need to specify a type signature. The monomorphism restriction would be useful in situations like: foo = expensiveOp bar How often will foo be evaluated? With the monomorphism restriction, at most once. Without it, well, that depends on the type of bar. With bar :: (?g :: Bool) => Int, foo also gets the (?g :: Bool) => context, and it's reevaluated whenever it's used; but you won't be able to tell that by looking at foo alone. Cheers, Wolfgang