Christopher Howard ch.howard at zoho.com
Sat Jun 18 05:30:16 UTC 2016

```Thanks, that was it!

On 06/17/2016 09:27 PM, Erik Rantapaa wrote:
> You are likely missing a `lift` when you call the random functions.
>
> Here is an example:
>
>
>
> walk :: RandomGen g => StateT (Float, Float) (Rand g) (Float, Float)
> walk = do (x, y) <- get
>           put (x + 1, y + 1)
>           get >>= return
>
> foo :: RandomGen g => StateT (Float,Float) (Rand g) ()
> foo = do
>   a <- lift \$ getRandomR (1,6)
>   b <- lift \$ getRandomR (4,10)
>   (x,y) <- get
>   put (x+a, y+b)
>
> test1 = do
>   g <- getStdGen
>   print \$ runRand (runStateT foo (0.0, 3.14)) g
>
> Because the State monad is encapsulating (transforming) the random
> monad, you have to `lift` operations in the random monad so that they
> become operations in the transformed monad.
>
> On Friday, June 17, 2016 at 11:22:57 PM UTC-5, Christopher Howard wrote:
>
>     Hi. I'm working through "Haskell Design Patterns" and got inspired to
>     try to create my first monad stack. What I really wanted though (not
>     shown in the book) was to combine State and Rand. I daresay I got
>     something to compile:
>
>     walk :: RandomGen g => StateT (Float, Float) (Rand g) (Float, Float)
>     walk = do (x, y) <- get
>               put (x + 1, y + 1)
>               get >>= return
>
>     However, the moment I try to insert a getRandomR or something in it, I
>     get an error
>
>     Could not deduce (MonadRandom (StateT (Float, Float) (Rand g)))
>           arising from a use of `getRandomR' <...snip...>
>     add an instance declaration for
>           (MonadRandom (StateT (Float, Float) (Rand g)))
>
>     I see there are instances
>
>     RandomGen g => MonadRandom (Rand g)
>
>     in Control.Monad.Random.Class, so I am not quite sure what is expected
>     of me.
>
>     --
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>
>     _______________________________________________