CODEWITHSUNDEEP
haskellmonadscontinuations
Sibi
Sibi

Reputation: 48644

Continuation Monad Working

I'm trying to wrap my head around this following piece of code:

ex2 = do
  a <- return 1
  b <- Cont (\fred -> fred 10)
  return $ a + b

ghci > runCont ex2 show
"11"

How is String "11" the output of the function ? I get that part that the show function converts the 10 to String, but then how does a + b take place since String numerals can't sum up ?

Upvotes: 1

Views: 500

Answers (1)

Chris Taylor
Chris Taylor

Reputation: 47382

The key is that a and b have type Int, and the conversion to a String with show is only done after a and b are summed.

If you need further convincing, ask yourself if you have any problem with

runCont ex2 id

or

runCont ex2 (\a -> a * a)

If you need even more convincing, note that the type definition and monad instance are

data Cont r a = Cont { runCont :: (a -> r) -> r }

instance Monad (Cont r) where
  return a = Cont ($a)

so you can squint a bit and convince yourself that your example is equivalent to

ex = do
  a <- return 1
  b <- return 10
  return (a + b)

which is just

ex = return (1 + 10)

Upvotes: 3

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