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-Random-Class.html#t:MonadRandom

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-Source.html

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Vanessa McHale
Functional Compiler Engineer | Chicago, IL

Website: www.iohk.io
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