It seems like you need your write-state to be a monoid in order to infer a `MonadRandom` instance. Based on your code, I'm guessing the compiler can't infer that. Have a look at https://hackage.haskell.org/package/MonadRandom-0.5.1.1/docs/Control-Monad-R... On 08/03/2018 07:31 AM, Sergiu Ivanov wrote:
Dear Café,
I would like to use random-fu to do some pseudo-random simulations for a given StdGen (so that I can run the same simulation multiple times, if needed).
The following works:
testState :: StdGen -> Int testState = evalState (sampleRVar $ uniform 1 10)
The following doesn't:
testRWS :: StdGen -> Int testRWS = fst . evalRWS (sampleRVar $ uniform 1 10) ()
I get
<interactive>:2:26: error: • No instance for (MonadRandom (RWST () b0 StdGen Identity)) arising from a use of ‘sampleRVar’ • In the first argument of ‘evalRWS’, namely ‘(sampleRVar $ uniform 1 10)’ In the second argument of ‘(.)’, namely ‘evalRWS (sampleRVar $ uniform 1 10) ()’ In the expression: fst . evalRWS (sampleRVar $ uniform 1 10) ()
Indeed, I do see a MonadRandom instance for StateT, but none for RWST [0].
Is there a reason to not have a MonadRandom instance for RWST?
Am I looking in the wrong place?
- Sergiu
[0] https://hackage.haskell.org/package/random-source-0.3.0.6/docs/Data-Random-S...
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