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Today's Topics:
1. Applicative on Tree (Mike Houghton)
2. Re: Applicative on Tree (Imants Cekusins)
3. Re: Applicative on Tree (Imants Cekusins)
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Message: 1
Date: Mon, 7 Sep 2015 20:59:18 +0100
From: Mike Houghton <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: [Haskell-beginners] Applicative on Tree
Message-ID: <[email protected]>
Content-Type: text/plain; charset=utf-8
Hi,
I?m looking at
data Tree a = Node a [Tree a] deriving (Show)
and trying to write the Applicative instance.
I have I think a correct Functor :
instance Functor (Tree) where
fmap f (Node x tr) = Node (f x) ( fmap (fmap f) tr)
but I?m well and truly stuck on Applicative :)
I have
instance Applicative (Tree) where
pure x = Node x []
(Node f []) <*> tr = fmap f tr
(Node f fs) <*> tr@(Node x xs) = ?????
I cant get rid of the question marks! :)
any pointers would be really appreciated!
Thanks
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Message: 2
Date: Tue, 8 Sep 2015 01:39:04 +0200
From: Imants Cekusins <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] Applicative on Tree
Message-ID:
<cap1qina55rwjye0qm-ksjahe5omrx88dnxf0f_eziz7r3-j...@mail.gmail.com>
Content-Type: text/plain; charset=UTF-8
> data Tree a = Node a [Tree a]
well here is something that builds and runs. Not sure if the monad laws apply.
watch out for indents!
{-# LANGUAGE InstanceSigs #-}
module TreeApp where
import Debug.Trace
import Data.Char
data Tree a = Node a [Tree a] deriving (Show)
instance Functor Tree where
fmap::(a -> b) -> Tree a -> Tree b
fmap f (Node x l0) = Node (f x) (fmap f <$> l0)
instance Applicative Tree where
pure::a -> Tree a
pure a = Node a []
(<*>)::Tree (a -> b) -> Tree a -> Tree b
(<*>) (Node f _) tra = f <$> tra
instance Monad Tree where
return::a -> Tree a
return a = pure a
(>>=)::Tree a -> (a -> Tree b) -> Tree b
(>>=) (Node x []) amb = amb x
(>>=) (Node x l0) amb = Node b (m1 <$> l0)
where (Node b _) = amb x
m1 ta = ta >>= amb
f::Char->Int
f = digitToInt
mb::Char->Tree Int
mb c = f <$> (pure c)
main::Char -> Char -> IO ()
main a b = print $ tc4 >>= mb
{-
do
print ti2
print ta3
tm4 <- tc4
-}
where tc1 = pure a
ti2 = f <$> tc1
ta3 = (Node f []) <*> tc1
tc4 = Node b [tc1]
------------------------------
Message: 3
Date: Tue, 8 Sep 2015 02:32:01 +0200
From: Imants Cekusins <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] Applicative on Tree
Message-ID:
<cap1qinbiw1rm_rvsdqizhpisdf-jczpgpdkkgxgadzfkzbp...@mail.gmail.com>
Content-Type: text/plain; charset=UTF-8
#2 .. this is probably correct:
{-# LANGUAGE InstanceSigs #-}
module TreeApp where
import Debug.Trace
import Data.Char
data Tree a = Node a [Tree a] deriving (Show)
instance Functor Tree where
fmap::(a -> b) -> Tree a -> Tree b
fmap f (Node x l0) = Node (f x) (fmap f <$> l0)
instance Applicative Tree where
pure::a -> Tree a
pure a = Node a []
(<*>)::Tree (a -> b) -> Tree a -> Tree b
(<*>) (Node f []) tra = f <$> tra
(<*>) tab@(Node f tf0) (Node x l0) = Node (f x) l1
where l1 = [fu <*> a | fu <- tf0 , a <- l0]
instance Monad Tree where
return::a -> Tree a
return a = pure a
(>>=)::Tree a -> (a -> Tree b) -> Tree b
(>>=) (Node x []) amb = amb x
(>>=) (Node x l0) amb = Node b (m1 <$> l0)
where (Node b _) = amb x
m1 ta = ta >>= amb
f1::Char->Int
f1 = digitToInt
mb::Char->Tree Int
mb c = f1 <$> (pure c)
main::Char -> Char -> IO ()
main a b = print $ tc4 >>= mb
where tc1 = Node a [Node b []]
ti2 = f1 <$> tc1
ta3 = (Node f1 []) <*> tc1
tc4 = Node b [tc1]
------------------------------
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