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Today's Topics:
1. ghci and randomRs (Jeff C. Britton)
2. Re: ghci and randomRs (Brandon Allbery)
3. Re: ghci and randomRs (David McBride)
----------------------------------------------------------------------
Message: 1
Date: Tue, 16 Sep 2014 02:01:30 +0000
From: "Jeff C. Britton" <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: [Haskell-beginners] ghci and randomRs
Message-ID:
<[email protected]>
Content-Type: text/plain; charset="us-ascii"
I am trying to use randomRs at the ghci prompt like so
Prelude System.Random> take 10 $ (randomRs (1, 6) newStdGen)
but I get the following error
<interactive>:16:12:
Could not deduce (RandomGen (IO StdGen))
arising from a use of `randomRs'
from the context (Random a, Num a)
bound by the inferred type of it :: (Random a, Num a) => [a]
at <interactive>:16:1-37
In the second argument of `($)', namely
`(randomRs (1, 6) newStdGen)'
In the expression: take 10 $ (randomRs (1, 6) newStdGen)
In an equation for `it': it = take 10 $ (randomRs (1, 6) newStdGen)
I have tried a variety of options, like wrapping it in a "do" or adding type
annotations.
Nothing seems to work.
-----Original Message-----
From: Beginners [mailto:[email protected]] On Behalf Of martin
Sent: Friday, September 12, 2014 10:05 AM
To: The Haskell-Beginners Mailing List - Discussion of primarily beginner-level
topics related to Haskell
Subject: Re: [Haskell-beginners] How to add a "method" to a record
Am 09/10/2014 08:50 PM, schrieb Corentin Dupont:
> If the field "label" can be deduced from "payload", I recommend not to
> include it in your structure, because that would be redundant.
>
> Here how you could write it:
>
> data Foo pl = Foo { payload :: pl}
>
> labelInt :: Foo Int -> String
> labelInt (Foo a) = "Int payload:" ++ (show a)
>
> labelString :: Foo String -> String
> labelString (Foo a) = "String payload" ++ a
>
> You are obliged to define two separate label function, because "Foo Int" and
> "Foo String" are two completly separate types.
This is exactly my problem: Someone will use this type an define the type of
pl. How can I know what type she'll use?
What I'd like to express is that whoever creates a concrete type should also
provide the proper label function.
>
> On Wed, Sep 10, 2014 at 2:06 PM, martin <[email protected]>
> wrote:
>
> Hello all
>
> if I have a record like
>
> data Foo pl = Foo {
> label :: String,
> payload :: pl
> }
>
> how can I create a similar type where I can populate label so it
> is not a plain string, but a function which
> operates on
> payload? Something like
>
> label (Foo pl) = show pl
>
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Message: 2
Date: Mon, 15 Sep 2014 22:05:52 -0400
From: Brandon Allbery <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] ghci and randomRs
Message-ID:
<cakfcl4xvyx_ppgakyqb47zrn8ydkxagkhpmugvsc6ykge_5...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
On Mon, Sep 15, 2014 at 10:01 PM, Jeff C. Britton <[email protected]> wrote:
> Prelude System.Random> take 10 $ (randomRs (1, 6) newStdGen)
I think you need to read http://www.vex.net/~trebla/haskell/IO.xhtml.
You're trying to operate directly on an IO "program" (newStdGen) as if it
were a value.
--
brandon s allbery kf8nh sine nomine associates
[email protected] [email protected]
unix, openafs, kerberos, infrastructure, xmonad http://sinenomine.net
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Message: 3
Date: Mon, 15 Sep 2014 22:27:02 -0400
From: David McBride <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] ghci and randomRs
Message-ID:
<CAN+Tr41=8h2pdJOBzePv+XjGHyw-FE5TgMMrB+DFwcfZ6dr=v...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
The simple answer is with do notation:
main = do
g <- newStdGen
print $ randomRs (1,2) g
Or without do notation, something like:
newStdGen >>= print . take 10 . randomRs (1,2)
On Mon, Sep 15, 2014 at 10:01 PM, Jeff C. Britton <[email protected]> wrote:
> I am trying to use randomRs at the ghci prompt like so
>
> Prelude System.Random> take 10 $ (randomRs (1, 6) newStdGen)
>
> but I get the following error
> <interactive>:16:12:
> Could not deduce (RandomGen (IO StdGen))
> arising from a use of `randomRs'
> from the context (Random a, Num a)
> bound by the inferred type of it :: (Random a, Num a) => [a]
> at <interactive>:16:1-37
> In the second argument of `($)', namely
> `(randomRs (1, 6) newStdGen)'
> In the expression: take 10 $ (randomRs (1, 6) newStdGen)
> In an equation for `it': it = take 10 $ (randomRs (1, 6) newStdGen)
>
>
> I have tried a variety of options, like wrapping it in a "do" or adding
> type annotations.
> Nothing seems to work.
>
>
>
> -----Original Message-----
> From: Beginners [mailto:[email protected]] On Behalf Of martin
> Sent: Friday, September 12, 2014 10:05 AM
> To: The Haskell-Beginners Mailing List - Discussion of primarily
> beginner-level topics related to Haskell
> Subject: Re: [Haskell-beginners] How to add a "method" to a record
>
> Am 09/10/2014 08:50 PM, schrieb Corentin Dupont:
> > If the field "label" can be deduced from "payload", I recommend not to
> > include it in your structure, because that would be redundant.
> >
> > Here how you could write it:
> >
> > data Foo pl = Foo { payload :: pl}
> >
> > labelInt :: Foo Int -> String
> > labelInt (Foo a) = "Int payload:" ++ (show a)
> >
> > labelString :: Foo String -> String
> > labelString (Foo a) = "String payload" ++ a
> >
> > You are obliged to define two separate label function, because "Foo Int"
> and "Foo String" are two completly separate types.
>
> This is exactly my problem: Someone will use this type an define the type
> of pl. How can I know what type she'll use?
> What I'd like to express is that whoever creates a concrete type should
> also provide the proper label function.
>
> >
> > On Wed, Sep 10, 2014 at 2:06 PM, martin <
> [email protected]> wrote:
> >
> > Hello all
> >
> > if I have a record like
> >
> > data Foo pl = Foo {
> > label :: String,
> > payload :: pl
> > }
> >
> > how can I create a similar type where I can populate label
> so it is not a plain string, but a function which
> > operates on
> > payload? Something like
> >
> > label (Foo pl) = show pl
> >
>
> _______________________________________________
> Beginners mailing list
> [email protected]
> http://www.haskell.org/mailman/listinfo/beginners
> _______________________________________________
> Beginners mailing list
> [email protected]
> http://www.haskell.org/mailman/listinfo/beginners
>
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