[Haskell-beginners] Temporary values with polymorphic types
Daniel Fischer
daniel.is.fischer at googlemail.com
Tue Feb 28 18:28:20 CET 2012
On Tuesday 28 February 2012, 17:53:10, Amy de Buitléir wrote:
> From this, I gather that I need to specify a type for the temporary
> graph "g"? I can arbitrarily pick an instance of the Graph class, but
> what can I put for the type parameter that it expects? It should be of
> the same type as the input array elements. This doesn't compile:
>
> ----- 8< -----
> import Data.Graph.Inductive.Graph ( labNodes, mkGraph )
> import qualified Data.Graph.Inductive.Tree as T ( Gr )
>
> doSomething ∷ [a] -> [a]
> doSomething xs = map snd $ labNodes g
> where xs' = zip [1..] xs
> g = mkGraph xs' [] :: T.Gr a Int
> ----- 8< -----
>
> amy3.hs:7:21:
> Couldn't match type `a' with `a2'
> `a' is a rigid type variable bound by
> the type signature for doSomething :: [a] -> [a] at
> amy3.hs:5:1 `a2' is a rigid type variable bound by
> an expression type signature: T.Gr a2 Int at amy3.hs:7:13
> Expected type: [Data.Graph.Inductive.Graph.LNode a1]
> Actual type: [(Int, a)]
> In the first argument of `mkGraph', namely `xs''
> In the expression: mkGraph xs' [] :: T.Gr a Int
> Failed, modules loaded: none.
The problem is that the 'a' in the type signature for the local g is a
fresh type variable, not the 'a' from the top level signature.
You can
a) bring the type variable into scope,
{-# LANGUAGE ScopedTypeVariables #-}
doSomething :: forall a. [a] -> [a]
doSomething xs = ...
where
xs' = zip [1 .. ] xs
g :: T.Gr a Int -- now it's the same a as in the top-level
g = mkGraph xs' []
b) use a type-restricted alias for mkGraph,
mkGraph' :: [(Int,a)] -> [??] -> T.Gr a Int
mkGraph' = mkGraph
and use mkGraph' in doSomething
There are probably more possibilities, but those are the only ones I found
without thinking.
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