Re: [Haskell-beginners] ghci and randomRs

Ok, I got it now. I was misunderstanding how the REPL was interacting with the IO Monad. I had once tried do { g <- newStdGen; take 10 $ randomRs (1,6) g } but I actually needed this do { g <- newStdGen; return . take 10 $ randomRs (1,6) g } Thanks, Jeff From: Beginners [mailto:beginners-bounces@haskell.org] On Behalf Of David McBride Sent: Monday, September 15, 2014 7:27 PM To: The Haskell-Beginners Mailing List - Discussion of primarily beginner-level topics related to Haskell Subject: Re: [Haskell-beginners] ghci and randomRs 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 mailto:jcb@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@haskell.orgmailto:beginners-bounces@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 mailto:martin.drautzburg@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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