问题
This code breaks when a type declaration for baz is added:
baz (x:y:_) = x == y
baz [_] = baz []
baz [] = False
A common explanation (see Why can't I declare the inferred type? for an example) is that it's because of polymorphic recursion.
But that explanation doesn't explain why the effect disappears with another polymorphically recursive example:
foo f (x:y:_) = f x y
foo f [_] = foo f []
foo f [] = False
It also doesn't explain why GHC thinks the recursion is monomorphic without type declaration.
Can the explanation of the example with reads in http://www.haskell.org/onlinereport/decls.html#sect4.5.5 be applied to my baz case?
I.e. adding a signature removes monomorphism restriction, and without the restriction an ambiguity of right-side [] appears, with an 'inherently ambigous' type of forall a . Eq a => [a]?
回答1:
The equations for baz are in one binding group, generalisation is done after the entire group has been typed. Without a type signature, that means baz is assumed to have a monotype, so the type of [] in the recursive call is given by that (look at ghc's -ddump-simpl output). With a type signature, the compiler is explicitly told that the function is polymorphic, so it can't assume the type of [] in the recursive call to be the same, hence it's ambiguous.
As John L said, in foo, the type is fixed by the occurrence of f - as long as f has a monotype. You can create the same ambiguity by giving f the same type as (==) (which requires Rank2Types),
{-# LANGUAGE Rank2Types #-}
foo :: Eq b => (forall a. Eq a => a -> a -> Bool) -> [b] -> Bool
foo f (x:y:_) = f x y
foo f[_] = foo f []
foo _ [] = False
That gives
Ambiguous type variable `b0' in the constraint:
(Eq b0) arising from a use of `foo'
Probable fix: add a type signature that fixes these type variable(s)
In the expression: foo f []
In an equation for `foo': foo f [_] = foo f []
回答2:
Your second example isn't polymorphically recursive. This is because the function f appears on both the LHS and RHS of the recursive definition. Also consider the type of foo, (a -> a -> Bool) -> [a] -> Bool. This fixes the list element type to be identical to the type of f's arguments. As a result, GHC can determine that the empty list on the RHS must have the same type as the input list.
I don't think that the reads example is applicable to the baz case, because GHC is able to compile baz with no type signature and the monomorphism restriction disabled. Therefore I expect that GHC's type algorithm has some other mechanism by which it removes the ambiguity.
来源：https://stackoverflow.com/questions/8270548/effects-of-monomorphism-restriction-on-type-class-constraints