* Francesco Mazzoli [2013-01-02 13:04:36+0100]

At Wed, 02 Jan 2013 12:32:53 +0100, Francesco Mazzoli wrote:

...

Hi list,

I am a bit puzzled by the behaviour exemplified by this code:

{-# LANGUAGE RankNTypes #-}

one :: (forall a. a -> a) -> b -> b one f = f

two = let f = flip one in f 'x' id three = (flip one :: b -> (forall a. a -> a) -> b) 'x' id four = flip one 'x' id

Try to guess if this code typechecks, and if not what’s the error.

While `two' and `three' are fine, GHC (7.4.1 and 7.6.1) complains about `four':

Line 8: 1 error(s), 0 warning(s)

Couldn't match expected type `forall a. a -> a' with actual type `a0 -> a0' In the third argument of `flip', namely `id' In the expression: flip one 'x' id In an equation for `four': four = flip one 'x' id

So for some reason the quantified variable in `id' gets instantiated before it should, and I have no idea why.

Any ideas?

Francesco

OK, I should have looked at the manual first. From http://www.haskell.org/ghc/docs/latest/html/users_guide/other-type-extension...: “For a lambda-bound or case-bound variable, x, either the programmer provides an explicit polymorphic type for x, or GHC's type inference will assume that x's type has no foralls in it.”. So there is a difference between let-bound things and the rest.

I don't see how this is relevant. GHC correctly infers the type of "flip one 'x'": *Main> :t flip one 'x' flip one 'x' :: (forall a. a -> a) -> Char But then somehow it fails to apply this to id. And there are no bound variables here that we should need to annotate. Roman

Your example doesn't work for the same reason the following doesn't work: id runST (<some st code>) It requires the inferencer to instantiate certain variables of id's type to polymorphic types based on runST (or flip's based on one), and then use that information to check <some st code> (id in your example) as a polymorphic type. At various times, GHC has had ad-hoc left-to-right behavior that made this work, but it no longer does. Right now, I believe it only has an ad-hoc check to make sure that: runST $ <some st code> works, and not much else. Note that even left-to-right behavior covers all cases, as you might have: f x y such that y requires x to be checked polymorphically in the same way. There are algorithms that can get this right in general, but it's a little tricky, and they're rather different than GHC's algorithm, so I don't know whether it's possible to make GHC behave correctly. The reason it works when you factor out or annotate "flip one 'x'" is that that is the eventual inferred type of the expression, and then it knows to expect the id to be polymorphic. But when it's all at once, we just have a chain of unifications relating things like: (forall a. a -> a) ~ beta ~ (alpha -> alpha), where beta is part of type checking flip, and alpha -> alpha is the instantiation of id's type with unification variables, because we didn't know that it was supposed to be a fully polymorphic use. And that unification fails. On Wed, Jan 2, 2013 at 8:21 AM, Francesco Mazzoli wrote:

At Wed, 2 Jan 2013 14:49:51 +0200, Roman Cheplyaka wrote:

...

I don't see how this is relevant.

Well, moving `flip one' in a let solves the problem, and The fact that let-bound variables are treated differently probably has a play here. I originally thought that this was because the quantifications will be all to the left in the let-bound variable while without a let-bound variable the types are used directly. However this doesn’t explain the behaviour I’m seeing.

...

GHC correctly infers the type of "flip one 'x'":

*Main> :t flip one 'x' flip one 'x' :: (forall a. a -> a) -> Char

But then somehow it fails to apply this to id. And there are no bound variables here that we should need to annotate.

Right. The weirdest thing is that annotating `flip one' (as in `three' in my code) or indeed `flip one 'x'' with the type that shows up in ghci makes things work:

five = (flip one 'x' :: (forall a. a -> a) -> Char) id

Francesco

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At Wed, 2 Jan 2013 11:20:46 -0500, Dan Doel wrote:

Your example doesn't work for the same reason the following doesn't work:

id runST (<some st code>)

It requires the inferencer to instantiate certain variables of id's type to polymorphic types based on runST (or flip's based on one), and then use that information to check <some st code> (id in your example) as a polymorphic type. At various times, GHC has had ad-hoc left-to-right behavior that made this work, but it no longer does. Right now, I believe it only has an ad-hoc check to make sure that:

runST $ <some st code>

works, and not much else. Note that even left-to-right behavior covers all cases, as you might have:

f x y

such that y requires x to be checked polymorphically in the same way. There are algorithms that can get this right in general, but it's a little tricky, and they're rather different than GHC's algorithm, so I don't know whether it's possible to make GHC behave correctly.

The reason it works when you factor out or annotate "flip one 'x'" is that that is the eventual inferred type of the expression, and then it knows to expect the id to be polymorphic. But when it's all at once, we just have a chain of unifications relating things like: (forall a. a -> a) ~ beta ~ (alpha -> alpha), where beta is part of type checking flip, and alpha -> alpha is the instantiation of id's type with unification variables, because we didn't know that it was supposed to be a fully polymorphic use. And that unification fails.

Hi Dan, Thanks a lot for the answer, one forgets that with HM you always replace the quantified variables immediately. However I am still confused on how GHC makes it work when I annotate or put things in separate variables. In other words, can you provide links or clarify how this procedure works: The reason it works when you factor out or annotate "flip one 'x'" is that that is the eventual inferred type of the expression, and then it knows to expect the id to be polymorphic. Thanks, Francesco

At Wed, 2 Jan 2013 13:35:24 -0500, Dan Doel wrote:

If you want to know the inner workings, you probably need to read the OutsideIn(X) paper.*

I'm not that familiar with the algorithm. But what happens is something like this.... When GHC goes to infer the type of 'f x' where it knows that f's argument is expected to be polymorphic, this triggers a different code path that will check that x can be given a type that is at least as general as is necessary for the argument.

However, "flip one 'x' id" gives flip a type like (alpha -> beta -> gamma) -> beta -> alpha -> gamma. Then, we probably get some constraints collected up like:

alpha ~ (forall a. a -> a) alpha ~ (delta -> delta)

That is, it does not compute the higher-rank type of "flip one 'x'" and then decide how the application of that to id should be checked; it decides how all the arguments should be checked based only on flip's type, and flip does not have a higher-rank type on its own. And solving the above constraints cannot trigger the alternate path.

However, when you factor out or annotate "flip one 'x'", it knows that it's applying something with a higher-rank type (whether because it inferred it separately, or you gave it), and that does trigger the alternate code path.

If that's still too vague, you'll have to refer to the paper.

-- Dan

* http://research.microsoft.com/en-us/um/people/simonpj/papers/constraints/jfp...

Thanks again for the answer. I understood more or less what was going on with the constraints - what I was wondering is how that alternate code path you cite works. I guess I’ll have to attack that epic paper sooner or later :). Francesco