[Haskell-beginners] ghci and randomRs

[David McBride]

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 <jcb at iteris.com> 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:beginners-bounces at haskell.org] 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 <
> martin.drautzburg at web.de> 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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