Sign Up

Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.

Have an account? Sign In

Have an account? Sign In Now

Sign In

Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.

Sign Up Here

Forgot Password?

Don't have account, Sign Up Here

Forgot Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Have an account? Sign In Now

You must login to ask a question.

Forgot Password?

Need An Account, Sign Up Here

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

Sign InSign Up

The Archive Base

The Archive Base Logo The Archive Base Logo

The Archive Base Navigation

  • SEARCH
  • Home
  • About Us
  • Blog
  • Contact Us
Search
Ask A Question

Mobile menu

Close
Ask a Question
  • Home
  • Add group
  • Groups page
  • Feed
  • User Profile
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Buy Points
  • Users
  • Help
  • Buy Theme
  • SEARCH
Home/ Questions/Q 6885421
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 27, 20262026-05-27T05:37:58+00:00 2026-05-27T05:37:58+00:00

Here are some pragmas and some imports: {-# LANGUAGE ScopedTypeVariables #-} import Control.Monad.ST import

  • 0

Here are some pragmas and some imports:

{-# LANGUAGE ScopedTypeVariables #-}

import Control.Monad.ST
import Data.Array.ST
import Data.Array

Now here’s my problem. The following code typechecks:

foo :: forall a. a -> [a]
foo x = elems $ runSTArray $ do
    newListArray (1,10) (replicate 10 x) :: ST s (STArray s Int a)

However, when I replace the $ with composition:

foo :: forall a. a -> [a]
foo x = elems . runSTArray $ do
    newListArray (1,10) (replicate 10 x) :: ST s (STArray s Int a)

I get this error:

Couldn't match expected type `forall s. ST s (STArray s i0 e0)'
            with actual type `ST s0 (STArray s0 Int a)'
In the expression:
    newListArray (1, 10) (replicate 10 x) :: ST s (STArray s Int a)
In the second argument of `($)', namely
  `do { newListArray (1, 10) (replicate 10 x) ::
          ST s (STArray s Int a) }'
In the expression:
      elems . runSTArray
  $ do { newListArray (1, 10) (replicate 10 x) ::
           ST s (STArray s Int a) }

What’s werid is, if I give the function composition its own name, then it typechecks again:

elemSTArray = elems . runSTArray

foo :: forall a. a -> [a]
foo x = elemSTArray $ do
    newListArray (1,10) (replicate 10 x) :: ST s (STArray s Int a)

I’m not sure what’s going on here. I would expect the second piece of code to typecheck nicely. And I don’t understand why it typechecks again if I give the composed function its own name.

This is a simplified version of some code that I had that broke when upgrading from GHC 6.2 to 7 and I’m trying to understand why this happens now. Thanks for helping!

  • 1 1 Answer
  • 0 Views
  • 0 Followers
  • 0
Share
  • Facebook
  • Report

Leave an answer
Cancel reply

You must login to add an answer.

Forgot Password?

Need An Account, Sign Up Here

1 Answer

  • Voted
  • Oldest
  • Recent
  • Random
  1. Editorial Team
    Editorial Team
    2026-05-27T05:37:58+00:00Added an answer on May 27, 2026 at 5:37 am

    As you already hint at in the title of your post, the problem has to do with runSTArray having a polymorphic type of rank 2.

    runSTArray :: Ix i => (forall s. ST s (STArray s i e)) -> Array i e
    

    With

    elems :: Ix i => Array i e -> [e]
    

    and

    ($) :: (a -> b) -> a -> b
    

    writing runSTArray $ ... means that the type variable a in the type schema of ($) needs to be instantiated with a polymorphic type rather than a monomorphic type. This requires so-called impredicative polymorphism. How GHC implements impredicative polymorphism is explained in the ICFP 2008 paper by Dimitrios Vytiniotis, Stephanie Weirich, and Simon Peyton Jones: FPH : First-class Polymorphism for Haskell. The bottom line is that while FPH often gives you the behaviour that you expect, typeability is sometimes not preserved under simple transformations like the ones you describe in your question: see Section 6.2 of the aforementioned paper.

    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Related Questions

Here is some code I could not get to format properly in markdown, this
Here's some Ruby code: puts %x{ pstree #{$$} } # never forks puts %x{
Here's some code I saw once. Can you see what's wrong with it? [updated]
Here is some XAML <HierarchicalDataTemplate DataType={x:Type data:FolderEntity} ItemsSource={Binding Path=FolderEntities,UpdateSourceTrigger=PropertyChanged}> <Label Content={Binding FolderName}/> </HierarchicalDataTemplate> <TreeView/>
Here is some simple code: DIR* pd = opendir(xxxx); struct dirent *cur; while (cur
Here's some code (full program follows later in the question): template <typename T> T
Here is some code I came accross for logging in but how would you
I've been looking for ways of solving this problem for quite some time now.
Here is some simple Perl to count the number of times a value occurs
Here's some background on what I'm trying to do: Open a serial port from

Explore

  • Home
  • Add group
  • Groups page
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Users
  • Help
  • SEARCH

Footer

© 2021 The Archive Base. All Rights Reserved
With Love by The Archive Base

Insert/edit link

Enter the destination URL

Or link to existing content

    No search term specified. Showing recent items. Search or use up and down arrow keys to select an item.