I've looked around at the various STRef examples out there, but still nothing I write myself using this will work. I'm trying to figure out how the s is escaping in really simple examples like x = runST $ return 1 y = runST $ do {r <- newSTRef 1; readSTRef r} Neither of these works in ghci - they both say <interactive>:1:0: Inferred type is less polymorphic than expected Quantified type variable `s' escapes Expected type: ST s a -> b Inferred type: (forall s1. ST s1 a) -> a In the first argument of `($)', namely `runST' In the definition of `it': ... I thought maybe I needed to replace 1 with (1 :: Int) so the state representation didn't force the type, but it still gives the same result. Can someone point me to the simplest possible runST example that actually works? Thanks! -- Chad Scherrer "Time flies like an arrow; fruit flies like a banana" -- Groucho Marx
Chad | x = runST $ return (1::Int) This code looks simple, but it isn't. Here are the types: runST :: forall a. (forall s. ST s a) -> a ($) :: forall b c. (b->c) -> b -> c return 1 :: forall s. ST s Int To typecheck, we must instantiate b with (forall s. ST s Int) c with Int In H-M that's impossible, because you can't instantiate a type variable (b) with a polytype (forall s. ST s Int). GHC will now let you do that (a rather recent change), but in this case it's hard to figure out that it should do so. Equally plausible is to instantiate b with (ST s' Int), where s' is a unification variable. One way to make this work is to look at $'s first argument first. Then it's clear how to instantiate b. Then look at the second argument. But if you look at the second argument first, matters are much less clear. GHC uses an algorithm that is insensitive to argument order, so it can't take advantage of the left-to-right bias of this example. It's unfortunate that such a simple-looking piece of code actually embodies a rather tricky typing problem! Of course there is no problem if you don' use the higher order function $. Use parens instead x = runST (return 1) Simon | -----Original Message----- | From: haskell-cafe-bounces@haskell.org [mailto:haskell-cafe-bounces@haskell.org] On Behalf Of | Chad Scherrer | Sent: 19 July 2006 23:02 | To: haskell-cafe@haskell.org | Subject: [Haskell-cafe] REALLY simple STRef examples | | I've looked around at the various STRef examples out there, but still | nothing I write myself using this will work. I'm trying to figure out | how the s is escaping in really simple examples like | | x = runST $ return 1 | | y = runST $ do {r <- newSTRef 1; readSTRef r} | | Neither of these works in ghci - they both say | | <interactive>:1:0: | Inferred type is less polymorphic than expected | Quantified type variable `s' escapes | Expected type: ST s a -> b | Inferred type: (forall s1. ST s1 a) -> a | In the first argument of `($)', namely `runST' | In the definition of `it': | ... | | I thought maybe I needed to replace 1 with (1 :: Int) so the state | representation didn't force the type, but it still gives the same | result. | | Can someone point me to the simplest possible runST example that | actually works? Thanks!
Thanks, Simon. I've begun putting together some text describing very
simple STRef examples, as Bulat suggested earlier. I think I know how
to make it work, but I'll still need to work on typing subtleties. I'm
headed to bed now, but I'll go through this in detail when I get a
chance to try to get my head around it.
-Chad
On 8/4/06, Simon Peyton-Jones
Chad
| x = runST $ return (1::Int)
This code looks simple, but it isn't. Here are the types:
runST :: forall a. (forall s. ST s a) -> a
($) :: forall b c. (b->c) -> b -> c
return 1 :: forall s. ST s Int
To typecheck, we must instantiate b with (forall s. ST s Int) c with Int
In H-M that's impossible, because you can't instantiate a type variable (b) with a polytype (forall s. ST s Int). GHC will now let you do that (a rather recent change), but in this case it's hard to figure out that it should do so. Equally plausible is to instantiate b with (ST s' Int), where s' is a unification variable.
One way to make this work is to look at $'s first argument first. Then it's clear how to instantiate b. Then look at the second argument. But if you look at the second argument first, matters are much less clear. GHC uses an algorithm that is insensitive to argument order, so it can't take advantage of the left-to-right bias of this example.
It's unfortunate that such a simple-looking piece of code actually embodies a rather tricky typing problem! Of course there is no problem if you don' use the higher order function $. Use parens instead
x = runST (return 1)
Simon
| -----Original Message----- | From: haskell-cafe-bounces@haskell.org [mailto:haskell-cafe-bounces@haskell.org] On Behalf Of | Chad Scherrer | Sent: 19 July 2006 23:02 | To: haskell-cafe@haskell.org | Subject: [Haskell-cafe] REALLY simple STRef examples | | I've looked around at the various STRef examples out there, but still | nothing I write myself using this will work. I'm trying to figure out | how the s is escaping in really simple examples like | | x = runST $ return 1 | | y = runST $ do {r <- newSTRef 1; readSTRef r} | | Neither of these works in ghci - they both say | | <interactive>:1:0: | Inferred type is less polymorphic than expected | Quantified type variable `s' escapes | Expected type: ST s a -> b | Inferred type: (forall s1. ST s1 a) -> a | In the first argument of `($)', namely `runST' | In the definition of `it': | ... | | I thought maybe I needed to replace 1 with (1 :: Int) so the state | representation didn't force the type, but it still gives the same | result. | | Can someone point me to the simplest possible runST example that | actually works? Thanks!
-- Chad Scherrer "Time flies like an arrow; fruit flies like a banana" -- Groucho Marx
participants (2)
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Chad Scherrer -
Simon Peyton-Jones