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Today's Topics:
1. Re: Functions as Applicatives (Olumide)
2. No instance for Show arising from a use in “main” level
(T. Andrea Moruno Rodriguez)
3. Re: No instance for Show arising from a use in “main”
level (T. Andrea Moruno Rodriguez)
4. putting together monadic actions (Dennis Raddle)
5. Re: putting together monadic actions (Sylvain Henry)
----------------------------------------------------------------------
Message: 1
Date: Tue, 23 Aug 2016 15:30:13 +0100
From: Olumide <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] Functions as Applicatives
Message-ID: <[email protected]>
Content-Type: text/plain; charset=utf-8; format=flowed
On 23/08/2016 12:39, Tony Morris wrote:
> All functions in Haskell always take one argument.
I know that. All functions accept one argument and return a value _or_
another function. Is f the latter type?
- Olumide
>
>
> On 23/08/16 21:28, Olumide wrote:
>> I must be missing something. I thought f accepts just one argument.
>>
>> - Olumide
>>
>> On 23/08/2016 00:54, Theodore Lief Gannon wrote:
>>> Yes, (g x) is the second argument to f. Consider the type signature:
>>>
>>> (<*>) :: Applicative f => f (a -> b) -> f a -> f b
>>>
>>> In this case, the type of f is ((->) r). Specialized to that type:
>>>
>>> (<*>) :: (r -> a -> b) -> (r -> a) -> (r -> b)
>>> f <*> g = \x -> f x (g x)
>>>
>>> Breaking down the pieces...
>>> f :: r -> a -> b
>>> g :: r -> a
>>> x :: r
>>> (g x) :: a
>>> (f x (g x)) :: b
>>>
>>> The example is made a bit confusing by tossing in an fmap. As far as the
>>> definition above is concerned, 'f' in the example is ((+) <$> (+3)) and
>>> that has to be resolved before looking at <*>.
>>>
>>>
>>> On Mon, Aug 22, 2016 at 9:07 AM, Olumide <[email protected]
>>> <mailto:[email protected]>> wrote:
>>>
>>> Hi List,
>>>
>>> I'm struggling to relate the definition of a function as a function
>>>
>>> instance Applicative ((->) r) where
>>> pure x = (\_ -> x)
>>> f <*> g = \x -> f x (g x)
>>>
>>> with the following expression
>>>
>>> ghci> :t (+) <$> (+3) <*> (*100)
>>> (+) <$> (+3) <*> (*100) :: (Num a) => a -> a
>>> ghci> (+) <$> (+3) <*> (*100) $ 5
>>> 508
>>>
>>> From chapter 11 of LYH http://goo.gl/7kl2TM .
>>>
>>> I understand the explanation in the book: "we're making a function
>>> that will use + on the results of (+3) and (*100) and return that.
>>> To demonstrate on a real example, when we did (+) <$> (+3) <*>
>>> (*100) $ 5, the 5 first got applied to (+3) and (*100), resulting in
>>> 8 and 500. Then, + gets called with 8 and 500, resulting in 508."
>>>
>>> The problem is that I can't relate that explanation with the
>>> definition of a function as an applicative; especially f <*> g = \x
>>> -> f x (g x) . Is (g x) the second argument to f?
>>>
>>> Regards,
>>>
>>> - Olumide
>>> _______________________________________________
>>> Beginners mailing list
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>>>
>>>
>>>
>>>
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>>
>> _______________________________________________
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>> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>
>
>
>
> _______________________________________________
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------------------------------
Message: 2
Date: Tue, 23 Aug 2016 20:10:34 -0400
From: "T. Andrea Moruno Rodriguez" <[email protected]>
To: [email protected]
Subject: [Haskell-beginners] No instance for Show arising from a use
in “main” level
Message-ID:
<CAFEhFEYyMvsGfU0vX=G-DaHtu-_o0_7zrb3FRQoJ=7g6uvu...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
I have a code that reads files and parses using UU.Parsing lib that returns
an abstract sintax tree and shows on the screen.
I received the error message "No instance for Show" in my functions
originated intokensParserToByteString and applyParser using parseIO (of
UU.Parsing lib) and inherited signatures until main. I fixed the signatures
but my problem is in the main function. I added the instance Show in the
signature but I have the next compilation error:
No instance for (Show (IO J2s)) arising from a use of ‘main’
In the expression: main
When checking the type of the IO action ‘main’
Some idea, about the problem?
Main module
-----------------------------------------------------------------------
{-# LANGUAGE FlexibleContexts #-}
module Main where
import UU.Parsing
...
import Content
main :: (Show (IO J2s)) => IO()
main = do f <- getLine
let command = test f
command
test :: (Show (IO J2s)) => String -> IO()
test "testparser" = testParser
Test module
-----------------------------------------------------------------------
{-# LANGUAGE FlexibleContexts #-}
module J2s.Parser.Test where
import Content
import J2s.Ast.Sintax
import J2s.Parser
import UU.Parsing
...
testParser :: (Show (IO J2s)) => IO()
testParser = (runSafeIO $ runProxy $ runEitherK $
contentsRecursive "path/of/my/tests" />/ handlerParser) ::
(Show (IO J2s)) => IO()
Content module
-----------------------------------------------------------------------
{-# LANGUAGE FlexibleContexts #-}
module Content where
import Control.Monad(forM, liftM)
import System.Directory (doesDirectoryExist, getDirectoryContents)
import System.FilePath ((</>), splitExtension, splitFileName)
import J2s.Parser
import J2s.Ast.Sintax
import UU.Parsing
import Control.Monad (when, unless)
import Control.Proxy
import Control.Proxy.Safe hiding (readFileS)
import J2s.Scanner.Token
import Text.Show
import UU.Parsing
contentsRecursive
:: (CheckP p)
=> FilePath -> () -> Producer (ExceptionP p) FilePath SafeIO ()
contentsRecursive path () = loop path
where
loop path = do
contents path () //> \newPath -> do
respond newPath
isDir <- tryIO $ doesDirectoryExist newPath
let isChild = not $ takeFileName newPath `elem` [".", ".."]
when (isDir && isChild) $ loop newPath
applyParser :: (Proxy p, Show (IO J2s)) => String -> Consumer p
B.ByteString IO ()
applyParser path = runIdentityP loop
where
loop = do
bs <- request ()
let sc = classify (initPos path) (B8.unpack bs)
lift $ B8.putStrLn (tokensParserToByteString sc)
tokensParserToByteString :: (Show (IO J2s)) => [Token] -> B.ByteString
tokensParserToByteString tokens = B8.pack(show (parseIO pJ2s tokens))
handlerParser :: (CheckP p, Show (IO J2s)) => FilePath -> Session
(ExceptionP p) SafeIO ()
handlerParser path = do
canRead <- tryIO $ fmap readable $ getPermissions path
isDir <- tryIO $ doesDirectoryExist path
isValidExtension <- tryIO $ evaluate ((snd (splitExtension path) ==
".java" || snd (splitExtension path) == ".mora") && (snd (splitFileName
path) /= "EncodeTest.java") && (snd (splitFileName path) /=
"T6302184.java") && (snd (splitFileName path) /= "Unmappable.java"))
when (not isDir && canRead && isValidExtension) $
(readFileSP 10240 path >-> try . applyParser) path
readFileSP
:: (CheckP p)
=> Int -> FilePath -> () -> Producer (ExceptionP p) B.ByteString SafeIO ()
readFileSP chunkSize path () =
bracket id (openFile path ReadMode) hClose $ \handle -> do
let loop = do
eof <- tryIO $ hIsEOF handle
unless eof $ do
bs <- tryIO $ B.hGetSome handle chunkSize
respond bs
loop
loop
--
Tatiana Andrea Moruno Rodriguez
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Message: 3
Date: Tue, 23 Aug 2016 20:44:04 -0400
From: "T. Andrea Moruno Rodriguez" <[email protected]>
To: Imants Cekusins <[email protected]>, [email protected]
Subject: Re: [Haskell-beginners] No instance for Show arising from a
use in “main” level
Message-ID:
<CAFEhFEa+18eTf_LXVR=ogf9nb+z7odhpgax1+vnxm6qosvw...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
Hi!
Yes.
Thanks I resolved my problem. The explanation is here:
http://stackoverflow.com/questions/39112380/no-instance-for-show-arising-from-a-use-in-main-level?noredirect=1#comment65572477_39112380
2016-08-23 20:23 GMT-04:00 Imants Cekusins <[email protected]>:
> Hello Tatiana,
>
> these signatures do not look good:
>
> main :: (Show (IO J2s)) => IO()
>
> test :: (Show (IO J2s)) => String -> IO()
>
> testParser :: (Show (IO J2s)) => IO()
>
>
> constraints are used to specify type variables. IO J2s looks like * (a
> fixed type).
>
> what errors do you see without these constraints?
>
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------------------------------
Message: 4
Date: Tue, 23 Aug 2016 18:00:34 -0700
From: Dennis Raddle <[email protected]>
To: Haskell Beginners <[email protected]>
Subject: [Haskell-beginners] putting together monadic actions
Message-ID:
<CAKxLvoo=-+eXc=ZKktv9+Ys+rKHfDFsRUPU+nPmtgSK=dvu...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
Is there a function foo that does
foo :: a -> [a -> m a] -> a
So
foo 3 [x,x,x] = return 3 >>= x >>= x >>= x
I don't think replicateM and sequence do this. At least I can't figure it
out.
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Message: 5
Date: Wed, 24 Aug 2016 03:33:01 +0200
From: Sylvain Henry <[email protected]>
To: [email protected]
Subject: Re: [Haskell-beginners] putting together monadic actions
Message-ID: <[email protected]>
Content-Type: text/plain; charset="utf-8"; Format="flowed"
Hi,
You can easily write your own:
> import Control.Monad
> import Data.List
> let foo = flip (foldl' (>=>) return)
> :t foo
foo :: (Foldable t, Monad m) => b -> t (b -> m b) -> m b
> let g x = return (x*2)
> foo 3 [g,g,g]
24
-Sylvain
On 24/08/2016 03:00, Dennis Raddle wrote:
> Is there a function foo that does
>
> foo :: a -> [a -> m a] -> a
>
> So
>
> foo 3 [x,x,x] = return 3 >>= x >>= x >>= x
>
> I don't think replicateM and sequence do this. At least I can't figure
> it out.
>
>
> _______________________________________________
> Beginners mailing list
> [email protected]
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