Thomas Davie
No, 'foo' does not literally gain the value of 'undefined'. In fact, foo gets no value at all, because the computation diverges before the pattern can be matched. /Semantically/ divergence is equivalent to 'undefined', but syntactically, 'foo' and 'undefined' are separate bindings, and their types are therefore not constrained to be equal.
Okay, but does this not force undefined to have the type Maybe a?
i.e. Just x = undefined File y = undefined
would cause a type error?
'undefined :: forall a . a' is a truly polymorphic value, in fact the only possible value of such an unconstrained type. In your example, it is instantiated twice, at different actual types, just as for instance the polymorphic function 'map' can be instantiated at several different actual types: map toUpper "hello" map (+1) [1,2,3] The only place where instantiating a polymorphic value at different types might cause a problem is in a recursive definition where the usage occurrence is at a different type to the defining occurrence, e.g. f x = f [x] Most functional languages do not permit this because it makes type inference undecidable. However Haskell allows it, provided that you give an explicit type signature so full inference is not required. Regards, Malcolm