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Today's Topics:
1. Bind parser returning wrong return type (Rohit Sharma)
2. Re: Bind parser returning wrong return type (Kim-Ee Yeoh)
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Message: 1
Date: Sat, 15 Nov 2014 20:10:55 +0800
From: Rohit Sharma <[email protected]>
To: [email protected]
Subject: [Haskell-beginners] Bind parser returning wrong return type
Message-ID:
<cabghn3bmem7meos96a1jhgd1vognh86fb6x3z9qab0iegwg...@mail.gmail.com>
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Hi All,
Sorry if this is a very lame question but i am a beginner and much
appreciate if some one could correct me?
I am reading haskell book and curious why the return type of the bind
operator look odd to me
For the given definitions
* type Parser a = String -> [(a, String)]*
* item :: Parser Char*
* item = \inp -> case inp of *
* [] -> []*
* (x:xs) -> [(x,xs)]*
* bind :: Parser a -> (a -> Parser b) -> Parser b*
* p `bind` f = \inp -> concat [ f x inp' | (x, inp') <- p inp]*
when I define z in GHCI as
* let z = item `bind` (\x -> (\y -> result (x,y))) "Rohit"*
the return type is
* >> :t z*
* z :: Parser ([Char], Char)*
Question:
(1) Shouldn't the return type of (Char, [Char])? looking at the list
comprehension, "(x, inp') <- p inp" should yield -> "('r', "ohit")". Next f
x inp' is left associative, so f x should yield character 'r' and pass to
the lambda that should return result tuple ('r', "ohit"), but why is it
that z type is ([Char], char) :: (x,y)
(2) How can i print the value of z in the above case on the ghci
Many thanks,
Rohit
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Message: 2
Date: Sat, 15 Nov 2014 22:08:49 +0700
From: Kim-Ee Yeoh <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] Bind parser returning wrong return
type
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<capy+zdrab2hkl+pf+rjdg-eyaxgwnjuv6hdsezkfz9mxwa9...@mail.gmail.com>
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Your questions aren't lame. They are common confusions from those who come
from a not-so-mathy background.
Your first question:
Shouldn't the return type be of (Char, [Char])?
suggests that you're confusing the type synonym of (Parser a), which is
String -> (a, String), with just the right-hand-side, (a, String).
Your "z" expression has type Parser ([Char], Char), which means the same
thing as String -> ((String, Char), String).
How did that happen?
Because
item `bind` (\x -> (\y -> result (x,y))) "Rohit"
is equivalent to
item `bind` ( (\x -> (\y -> result (x,y))) "Rohit" )
as those last two expressions go together by the parsing rules.
So what you actually have is
item `bind` (\y -> result ("Rohit",y))
The first argument to bind has type Parser Char, the second argument (a ->
Parser (String,a)).
The result is exactly what's expected: Parser (String, Char), i.e. String
-> ((String, Char), String).
p.s. This might be what you're looking for: Try evaluating
item "Rohit"
in the repl.
-- Kim-Ee
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