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:
import Control.Monad.State import Control.Monad.Random
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
MonadRandom m => MonadRandom (StateT s m) 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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