My completely off-the-cuff guess is that a a b b isn't considered more or less specific than (x -> a) ar (x -> b) br since they both apply some constraint on the types. For example, it's not immediately clear that the first instance can't be used for (x -> a) (x -> a) (x -> b) (x -> b) Whereas when you say a ar b br the type (x -> a) ar (x -> b) br is strictly more specific, so the overlapping instance can be chosen. Remember instance selection is done entirely via the instance head, so instance X a a is not the same as instance (a ~ b) => X a b The first case supplies an instance for any two equal types, and the second case supplies an instance for *any two types*, then throws an error if the compiler can't prove that the two types are equal. For example, without overlapping instances, you can write class X a b where foo :: a -> b instance X a a where foo = id instance X Int Bool where foo = (== 0) But if instead you specify instance (a ~ b) => X a b where foo = id you can't specify the Int Bool instance without overlap. -- ryan On Mon, Jul 30, 2012 at 12:32 PM, Artyom Kazak <artyom.kazak@gmail.com>wrote:

Hello,

I have accidentally written my version of polyvariadic composition combinator, `mcomp`. It differs from Oleg’s version ( http://okmij.org/ftp/Haskell/**polyvariadic.html#polyvar-comp<http://okmij.org/ftp/Haskell/polyvariadic.html#polyvar-comp>) in three aspects: a) it is simpler, b) it works without enumerating basic cases (all existing types, in other words), and c) it needs more type extensions.

{-# LANGUAGE

...

MultiParamTypeClasses , FunctionalDependencies , FlexibleInstances , UndecidableInstances , TypeFamilies , OverlappingInstances #-}

class Mcomp a ar b br | a br -> b where mcomp :: a -> (ar -> br) -> b

instance (a ~ ar, b ~ br) => Mcomp a ar b br where mcomp a f = f a

instance (Mcomp a ar b br) => Mcomp (x -> a) ar (x -> b) br where mcomp a f = \x -> mcomp (a x) f

My question is: why doesn’t it work when I replace

instance (a ~ ar, b ~ br) => Mcomp a ar b br

with

instance Mcomp a a b b

? I thought that equal letters mean equal types…

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