Type inference shows (Num (a -> a), Num a)

Defining f and g as:

f x = x + 1 g x y = x + y

Entering :t f . g gives: f . g :: (Num (a -> a), Num a) => a -> a -> a

Formally, it is correct, since Num (a -> a) is never going to be a number. Though me and my tutor thought it might be a bug, and f . g shouldn't be recognised as a well-formed expression.

Trac metadata

Trac field	Value
Version	8.0.1
Type	Bug
TypeOfFailure	OtherFailure
Priority	normal
Resolution	Unresolved
Component	Compiler (Type checker)
Test case
Differential revisions
BlockedBy
Related
Blocking
CC	lo@cs.ox.ac.uk
Operating system
Architecture

changed weight to 5

added Tbug Trac import typechecker labels

closed

Num (a -> a) is never going to be a number.

It could be. Someone just has to give an instance Num (a -> b). The compiler cannot assume that no such instance will ever be around.

It seems to me that the compiler behaviour is ok and desired as is.

Trac metadata

Trac field	Value
Resolution	Unresolved → ResolvedInvalid

added Pnormal label