haskell - How do bind, put and return interact here? -

given following code

rolldie :: generatorstate int rolldie = generator <-              let (value, newgenerator) = randomr (1,6) generator              put newgenerator              return value 

i know can translate as:

rolldie2 :: generatorstate int rolldie2 = >>= \generator ->let (value, newgenerator) = randomr(1,6) generator                                 in put newgenerator >> return value 

i tested both functions , without put newgenerator >>, , produce different results. question why? put functions pure, , operator (>>) means return value should unaffected prior results.

when test both functions with same initial state, same answer:

λ> evalstate rolldie (mkstdgen 0) 6 λ> evalstate rolldie2 (mkstdgen 0) 6 

i suspect you're not using same state both tests. how testing functions?

here's example state (i.e. random number generator) gets modified:

test :: generatorstate (int, int) test =   <- rolldie -- modifies state!   b <- rolldie2 -- gets different state   return (a, b)  runtest :: io () runtest =   g <- getstdgen   let (a, b) = evalstate test g   print   print b 

as can see, when run 2 different answers.

λ> runtest 4 2