Send Beginners mailing list submissions to
[email protected]
To subscribe or unsubscribe via the World Wide Web, visit
http://www.haskell.org/mailman/listinfo/beginners
or, via email, send a message with subject or body 'help' to
[email protected]
You can reach the person managing the list at
[email protected]
When replying, please edit your Subject line so it is more specific
than "Re: Contents of Beginners digest..."
Today's Topics:
1. A question about an infinite type (Costello, Roger L.)
2. Temporary values with polymorphic types (Amy de Buitl?ir)
3. Re: Temporary values with polymorphic types (Daniel Fischer)
----------------------------------------------------------------------
Message: 1
Date: Tue, 28 Feb 2012 12:52:19 +0000
From: "Costello, Roger L." <[email protected]>
Subject: [Haskell-beginners] A question about an infinite type
To: "[email protected]" <[email protected]>
Message-ID:
<[email protected]>
Content-Type: text/plain; charset="us-ascii"
Hi Folks,
Here is an interesting phenomena:
let f = (\x y -> x y x)
let p = (\x y -> 1)
Now let's evaluate f p 1
f p 1 = (\x y -> x y x) p 1 -- by replacing f with its
definition
= p 1 p -- by substituting x
with p and y with 1
= (\x y -> 1) 1 p -- by replacing the first p
with its definition
= 1 -- the function
returns 1 regardless of its arguments
However, Haskell will not proceed with that evaluation because it will first
determine the
type signature of f and judge it to be an infinite type.
Conversely, if f is omitted and this expression
p 1 p
is evaluated then the result 1 is generated.
Why does f p 1 (which evaluates to p 1 p) fail whereas p 1 p succeeds?
What lesson should I learn from this example?
/Roger
------------------------------
Message: 2
Date: Tue, 28 Feb 2012 16:53:10 +0000 (UTC)
From: Amy de Buitl?ir <[email protected]>
Subject: [Haskell-beginners] Temporary values with polymorphic types
To: [email protected]
Message-ID: <[email protected]>
Content-Type: text/plain; charset=utf-8
I'm trying to write a function that builds a temporary graph, performs some
operation on the graph, and then returns the result of the operation. The graph
isn't returned. Here's an example. (The real function is much more complicated.)
----- 8< -----
import Data.Graph.Inductive.Graph ( labNodes, mkGraph )
doSomething ? [a] -> [a]
doSomething xs = map snd $ labNodes $ mkGraph xs' []
where xs' = zip [1..] xs
----- 8< -----
When I load this in GHCI, I get:
amy3.hs:4:39:
Ambiguous type variable `gr0' in the constraint:
(Data.Graph.Inductive.Graph.Graph
gr0) arising from a use of `mkGraph'
Probable fix: add a type signature that fixes these type variable(s)
In the second argument of `($)', namely `mkGraph xs' []'
In the second argument of `($)', namely `labNodes $ mkGraph xs' []'
In the expression: map snd $ labNodes $ mkGraph xs' []
Failed, modules loaded: none.
>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.
Thank you in advance for any advice.
------------------------------
Message: 3
Date: Tue, 28 Feb 2012 18:28:20 +0100
From: Daniel Fischer <[email protected]>
Subject: Re: [Haskell-beginners] Temporary values with polymorphic
types
To: [email protected]
Cc: Amy de Buitl?ir <[email protected]>
Message-ID: <[email protected]>
Content-Type: Text/Plain; charset="utf-8"
On Tuesday 28 February 2012, 17:53:10, Amy de Buitl?ir 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.
------------------------------
_______________________________________________
Beginners mailing list
[email protected]
http://www.haskell.org/mailman/listinfo/beginners
End of Beginners Digest, Vol 44, Issue 28
*****************************************
