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Today's Topics:
1. Re: Could not get parser ready (David McBride)
2. Re: Could not get parser ready (Marcus Manning)
----------------------------------------------------------------------
Message: 1
Date: Mon, 6 Nov 2017 08:29:38 -0500
From: David McBride <toa...@gmail.com>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <beginners@haskell.org>
Subject: Re: [Haskell-beginners] Could not get parser ready
Message-ID:
<can+tr40lstbonvqeudc5thkckh2t0yfyqi5oqx6heif06ek...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
The problem is that in p you are using do notation for bind, which uses the
Monad instance for bind (>>=) for ((->) String), because Parser is a type
alias for (String -> [(a, String)]. But then you are using your own return
which is not of the same type as Monad's return would be. The way you have
it now your return considers the loose `a` as the Monad parameter, but do
notation considers it [(a, String)] instead.
You can either a) write your own bind that conforms to your type.
bind :: Parser a -> (a -> Parser b) -> Parser b
bind = undefined
p' :: Parser (Char, Char)
p' = item `bind` \x -> item `bind` \_ -> item `bind` \y -> Foo.return (x, y)
Or you can use the Monad return instead. But the type is not what you
expect.
p :: String -> ([(Char, String)], [(Char, String)])
p = do x <- item
item
y <- item
Prelude.return (x,y)
On Mon, Nov 6, 2017 at 3:15 AM, Marcus Manning <icons...@gmail.com> wrote:
> type Parser a = String → [(a,String)]
>
> item :: Parser Char
> item = λinp → case inp of
> [] → []
> (x:xs) → [(x,xs)]
> failure :: Parser a
> failure = λinp → []
>
> return :: a → Parser a
> return v = λinp → [(v,inp)]
>
> (+++) :: Parser a → Parser a → Parser a
> p +++ q = λinp → case p inp of
> [] → q inp
> [(v,out)] → [(v,out)]
>
> parse :: Parser a → String → [(a,String)]
> parse p inp = p inp
>
> p :: Parser (Char,Char)
> p = do x ← item
> item
> y ← item
> return (x,y)
>
> It is described in pages 189-216 in [1].
>
> [1] https://userpages.uni-koblenz.de/~laemmel/paradigms1011/resources/pdf/
> haskell.pdf
>
> I assume the bind operator (==>) was overwritten by
>
> (>>=) :: Parser a → (a → Parser b) → Parser b p
> >>= f = λinp → case parse p inp of
> [ ] → [ ]
> [ (v, out) ] → parse (f v) out
>
> in order to manipulate the do expr to make the p function work, right?
>
> 2017-11-05 21:56 GMT+01:00 Tobias Brandt <to...@uni-bremen.de>:
>
>> Hey,
>>
>> can you show us your Parser definition?
>>
>> Cheers,
>> Tobias
>>
>> ----- Nachricht von Marcus Manning <icons...@gmail.com> ---------
>> Datum: Sun, 5 Nov 2017 18:51:57 +0100
>> Von: Marcus Manning <icons...@gmail.com>
>> Antwort an: The Haskell-Beginners Mailing List - Discussion of primarily
>> beginner-level topics related to Haskell <beginners@haskell.org>
>> Betreff: [Haskell-beginners] Could not get parser ready
>> An: beginners@haskell.org
>>
>> Hello,
>>
>> I follow the instructions of script [1] in order to set up a parser
>> functionality. But I' get into problems at page 202 with the code:
>>
>> p :: Parser (Char,Char)
>> p = do
>> x ← item
>> item
>> y ← item
>> return (x,y)
>>
>>
>> ghci and ghc throw errors:
>> Prelude> let p:: Parser (Char,Char); p = do {x <- item; item; y <- item;
>> return (x,y)}
>>
>> <interactive>:10:65: error:
>> • Couldn't match type ‘[(Char, String)]’ with ‘Char’
>> Expected type: String -> [((Char, Char), String)]
>> Actual type: Parser ([(Char, String)], [(Char, String)])
>> • In a stmt of a 'do' block: return (x, y)
>> In the expression:
>> do x <- item
>> item
>> y <- item
>> return (x, y)
>> In an equation for ‘p’:
>> p = do x <- item
>> item
>> y <- item
>> ....
>> Did the semantics of do expr changed?
>>
>> [1] https://userpages.uni-koblenz.de/~laemmel/paradigms1011/reso
>> urces/pdf/haskell.pdf
>>
>> Cheers,
>>
>> iconfly.
>> _______________________________________________
>> Beginners mailing list
>> Beginners@haskell.orghttp://mail.haskell.org/cgi-bin/mailman
>> /listinfo/beginners
>>
>>
>>
>>
>> ----- Ende der Nachricht von Marcus Manning <icons...@gmail.com> -----
>>
>>
>> _______________________________________________
>> Beginners mailing list
>> Beginners@haskell.org
>> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>>
>>
>
> _______________________________________________
> Beginners mailing list
> Beginners@haskell.org
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>
>
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Message: 2
Date: Mon, 6 Nov 2017 16:48:04 +0100
From: Marcus Manning <icons...@gmail.com>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <beginners@haskell.org>
Subject: Re: [Haskell-beginners] Could not get parser ready
Message-ID:
<cahfjoxsf5zz28ehh78arrkfmjb8ygytvoew4vz97quofzwo...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
p' :: Parser (Char, Char)
p' = item `bind` \x -> item `bind` \_ -> item `bind` \y -> return (x, y)
with
bind:: Parser a -> (a -> Parser b) -> Parser b
bind p f = \ inp -> case parse p inp of
[] -> []
[(v,out)] -> parse (f v) out
works like a charm.
Another alternatiev would to change the Parser definition to the one given
in [1].
Thanks.
Cheers,
iconfly
[1] http://dev.stephendiehl.com/fun/002_parsers.html
2017-11-06 14:29 GMT+01:00 David McBride <toa...@gmail.com>:
> The problem is that in p you are using do notation for bind, which uses
> the Monad instance for bind (>>=) for ((->) String), because Parser is a
> type alias for (String -> [(a, String)]. But then you are using your own
> return which is not of the same type as Monad's return would be. The way
> you have it now your return considers the loose `a` as the Monad parameter,
> but do notation considers it [(a, String)] instead.
>
> You can either a) write your own bind that conforms to your type.
>
> bind :: Parser a -> (a -> Parser b) -> Parser b
> bind = undefined
>
> p' :: Parser (Char, Char)
> p' = item `bind` \x -> item `bind` \_ -> item `bind` \y -> Foo.return (x,
> y)
>
> Or you can use the Monad return instead. But the type is not what you
> expect.
>
> p :: String -> ([(Char, String)], [(Char, String)])
> p = do x <- item
> item
> y <- item
> Prelude.return (x,y)
>
>
>
>
>
> On Mon, Nov 6, 2017 at 3:15 AM, Marcus Manning <icons...@gmail.com> wrote:
>
>> type Parser a = String → [(a,String)]
>>
>> item :: Parser Char
>> item = λinp → case inp of
>> [] → []
>> (x:xs) → [(x,xs)]
>> failure :: Parser a
>> failure = λinp → []
>>
>> return :: a → Parser a
>> return v = λinp → [(v,inp)]
>>
>> (+++) :: Parser a → Parser a → Parser a
>> p +++ q = λinp → case p inp of
>> [] → q inp
>> [(v,out)] → [(v,out)]
>>
>> parse :: Parser a → String → [(a,String)]
>> parse p inp = p inp
>>
>> p :: Parser (Char,Char)
>> p = do x ← item
>> item
>> y ← item
>> return (x,y)
>>
>> It is described in pages 189-216 in [1].
>>
>> [1] https://userpages.uni-koblenz.de/~laemmel/paradigms1011/
>> resources/pdf/haskell.pdf
>>
>> I assume the bind operator (==>) was overwritten by
>>
>> (>>=) :: Parser a → (a → Parser b) → Parser b p
>> >>= f = λinp → case parse p inp of
>> [ ] → [ ]
>> [ (v, out) ] → parse (f v) out
>>
>> in order to manipulate the do expr to make the p function work, right?
>>
>> 2017-11-05 21:56 GMT+01:00 Tobias Brandt <to...@uni-bremen.de>:
>>
>>> Hey,
>>>
>>> can you show us your Parser definition?
>>>
>>> Cheers,
>>> Tobias
>>>
>>> ----- Nachricht von Marcus Manning <icons...@gmail.com> ---------
>>> Datum: Sun, 5 Nov 2017 18:51:57 +0100
>>> Von: Marcus Manning <icons...@gmail.com>
>>> Antwort an: The Haskell-Beginners Mailing List - Discussion of primarily
>>> beginner-level topics related to Haskell <beginners@haskell.org>
>>> Betreff: [Haskell-beginners] Could not get parser ready
>>> An: beginners@haskell.org
>>>
>>> Hello,
>>>
>>> I follow the instructions of script [1] in order to set up a parser
>>> functionality. But I' get into problems at page 202 with the code:
>>>
>>> p :: Parser (Char,Char)
>>> p = do
>>> x ← item
>>> item
>>> y ← item
>>> return (x,y)
>>>
>>>
>>> ghci and ghc throw errors:
>>> Prelude> let p:: Parser (Char,Char); p = do {x <- item; item; y <- item;
>>> return (x,y)}
>>>
>>> <interactive>:10:65: error:
>>> • Couldn't match type ‘[(Char, String)]’ with ‘Char’
>>> Expected type: String -> [((Char, Char), String)]
>>> Actual type: Parser ([(Char, String)], [(Char, String)])
>>> • In a stmt of a 'do' block: return (x, y)
>>> In the expression:
>>> do x <- item
>>> item
>>> y <- item
>>> return (x, y)
>>> In an equation for ‘p’:
>>> p = do x <- item
>>> item
>>> y <- item
>>> ....
>>> Did the semantics of do expr changed?
>>>
>>> [1] https://userpages.uni-koblenz.de/~laemmel/paradigms1011/reso
>>> urces/pdf/haskell.pdf
>>>
>>> Cheers,
>>>
>>> iconfly.
>>> _______________________________________________
>>> Beginners mailing list
>>> Beginners@haskell.orghttp://mail.haskell.org/cgi-bin/mailman
>>> /listinfo/beginners
>>>
>>>
>>>
>>>
>>> ----- Ende der Nachricht von Marcus Manning <icons...@gmail.com> -----
>>>
>>>
>>> _______________________________________________
>>> Beginners mailing list
>>> Beginners@haskell.org
>>> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>>>
>>>
>>
>> _______________________________________________
>> Beginners mailing list
>> Beginners@haskell.org
>> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>>
>>
>
> _______________________________________________
> Beginners mailing list
> Beginners@haskell.org
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>
>
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