Trouble with types
Hi, I tried this code: ----------------------- f, g :: a -> a (f, g) = (id, id) ----------------------- Hugs: OK GHC: Couldn't match expected type `forall a. a -> a' against inferred type `a -> a' In the expression: id In the expression: (id, id) In a pattern binding: (f, g) = (id, id) What does mean this error message? And what of them (Hugs, GHC) is correct? Thanks Vladimir
Am Montag 01 Juni 2009 14:44:37 schrieb Vladimir Reshetnikov:
Hi,
I tried this code:
----------------------- f, g :: a -> a (f, g) = (id, id) -----------------------
Hugs: OK
GHC: Couldn't match expected type `forall a. a -> a' against inferred type `a -> a' In the expression: id In the expression: (id, id) In a pattern binding: (f, g) = (id, id)
What does mean this error message? And what of them (Hugs, GHC) is correct?
http://www.haskell.org/ghc/docs/latest/html/users_guide/bugs-and-infelicitie... Section 12.1.1.4, Declarations and bindings GHC's typechecker makes all pattern bindings monomorphic by default; this behaviour can be disabled with -XNoMonoPatBinds. See Section 7.1, “Language options”. Hugs is correct, it's a known infelicity in GHC which can be disabled.
Thanks Vladimir
Hi Daniel,
Could you please explain what does mean 'monomorphic' in this context?
I thought that all type variables in Haskell are implicitly
universally quantified, so (a -> a) is the same type as (forall a. a
-> a)
Thank you,
Vladimir
On 6/1/09, Daniel Fischer
Am Montag 01 Juni 2009 14:44:37 schrieb Vladimir Reshetnikov:
Hi,
I tried this code:
----------------------- f, g :: a -> a (f, g) = (id, id) -----------------------
Hugs: OK
GHC: Couldn't match expected type `forall a. a -> a' against inferred type `a -> a' In the expression: id In the expression: (id, id) In a pattern binding: (f, g) = (id, id)
What does mean this error message? And what of them (Hugs, GHC) is correct?
http://www.haskell.org/ghc/docs/latest/html/users_guide/bugs-and-infelicitie... Section 12.1.1.4, Declarations and bindings
GHC's typechecker makes all pattern bindings monomorphic by default; this behaviour can be disabled with -XNoMonoPatBinds. See Section 7.1, “Language options”.
Hugs is correct, it's a known infelicity in GHC which can be disabled.
Thanks Vladimir
_______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Vladimir Reshetnikov wrote:
Hi Daniel,
Could you please explain what does mean 'monomorphic' in this context? I thought that all type variables in Haskell are implicitly universally quantified, so (a -> a) is the same type as (forall a. a -> a)
At the top level (i.e. definition level), yes. However, each use site this polymorphism may be restricted down (in particular, to the point of monomorphism). In a syntax more like Core, the definition of the identity function is, id :: forall a. a -> a id @a (x :: a) = x Where the @ is syntax for reifying types or for capital-lambda application (whichever interpretation you prefer). In this syntax it's clear to see that |id| (of type forall a. a -> a) is quite different than any particular |id @a| (of type a->a for the particular @a). So in your example, there's a big difference between these definitions, (f,g) = (id,id) (f', g') @a = (id @a, id @a) The latter one is polymorphic, but it has interesting type sharing going on which precludes giving universally quantified types to f' and g' (the universality is for the pair (f',g') and so the types of f' and g' must covary). Whereas the former has both fields of the tuple being polymorphic (independently). Both interpretations of the original code are legitimate in theory, but the former is much easier to work with and reason about. It also allows for tupling definitions as a way of defining local name scope blocks, without side effects on types. -- Live well, ~wren
participants (3)
-
Daniel Fischer -
Vladimir Reshetnikov -
wren ng thornton