Creating a mutable Data.Vector in Haskell

Question

I wish to create a mutable vector using Data.Vector.Generic.Mutable.new. I have found examples that create a mutable vector by thawing a pure vector, but that's not what I wish to do.

Here is one of many failed attempts:

import Control.Monad.Primitive
import qualified Data.Vector.Generic.Mutable as GM

main = do
  v <- (GM.new 10) :: (GM.MVector v a) => IO (v RealWorld a)
  GM.write v 0 (3::Int)
  x <- GM.read v 0
  putStrLn $ show x

giving me the error

No instance for (GM.MVector v0 Int)
  arising from an expression type signature
Possible fix: add an instance declaration for (GM.MVector v0 Int)

I tried variations based on the Haskell Vector tutorial with no luck.

I would also welcome suggestion on cleaner ways to construct the vector. The reference to RealWorld seems ugly to me.

cdk · Accepted Answer · 2014-03-30 14:54:40Z

5

The GM.MVector v a constaint is ambigous in v. In other words, from the type information you've given GHC, it still can't figure out what specific instance of GM.MVector you want it to use. For a mutable vector of Int use Data.Vector.Unboxed.Mutable.

import qualified Data.Vector.Unboxed.Mutable as M

main = do
    v <- M.new 10
    M.write v 0 (3 :: Int)
    x <- M.read v 0
    print x

edited Mar 30, 2014 at 14:54

answered Mar 30, 2014 at 5:20

cdk

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Comments

ErikR · Accepted Answer · 2014-03-30 05:09:38Z

2

I think the problem is that you have to give v a concrete type -- like this:

import Control.Monad.Primitive
import qualified Data.Vector.Mutable as V
import qualified Data.Vector.Generic.Mutable as GM

main = do
  v <- GM.new 10 :: IO (V.MVector RealWorld Int)
  GM.write v 0 (3::Int)
  x <- GM.read v 0
  putStrLn $ show x

answered Mar 30, 2014 at 5:09

ErikR

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1 Comment

chris Over a year ago

The close correspondence with my code helped me improve my type declaration skills.

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