How does forever works ?
From what I understand, this code
main = forever $ do putStrLn "OK !" is equivalent to this one : main = do putStrLn "OK !" main In the second case, it's a simple recursion, so far so good ... but when I look at the implementation of "forever" i can't wrap my head around : http://hackage.haskell.org/package/base-4.11.1.0/docs/src/Control.Monad.html... forever http://hackage.haskell.org/package/base-4.11.1.0/docs/src/Control.Monad.html... a http://hackage.haskell.org/package/base-4.11.1.0/docs/src/Control.Monad.html... = let a' http://hackage.haskell.org/package/base-4.11.1.0/docs/src/Control.Monad.html... = a http://hackage.haskell.org/package/base-4.11.1.0/docs/src/Control.Monad.html... *> http://hackage.haskell.org/package/base-4.11.1.0/docs/src/GHC.Base.html#%2A%... a' http://hackage.haskell.org/package/base-4.11.1.0/docs/src/Control.Monad.html... in a' http://hackage.haskell.org/package/base-4.11.1.0/docs/src/Control.Monad.html... How does this works ? How does this make an infinite loop ? I understand that *> discard his right argument but it doesn't help me understand how forever implement an infinite loop ...
Hello Olivier, On Mon, Jul 09, 2018 at 01:31:07PM +0200, Olivier Revollat wrote:
In the second case, it's a simple recursion, so far so good ... but when I look at the implementation of "forever" i can't wrap my head around :
http://hackage.haskell.org/package/base-4.11.1.0/docs/src/Control.Monad.html...
forever a = let a' = a *> a' in a' `*>` is the same as `>>` (if you have ever met `>>`), with the difference that the latter works on monads only. In do notation, we could write forever :: Monad m => m a -> m b forever a = let a' = do a a' in do a' Which should be easier to get: we `do` a'; a' is nothing but a and a', so we do a and then a', which is nothing but a and a', etc. Does this explanation feel right? As soon as you can, familiarise yourself what the do notation desugars to: in my opinion plain operators are clearer -F
Hey I think it's a bit clearer now .... need to meditate on this :) Thanks !
Le lun. 9 juil. 2018 à 13:55, Francesco Ariis
Hello Olivier,
On Mon, Jul 09, 2018 at 01:31:07PM +0200, Olivier Revollat wrote:
In the second case, it's a simple recursion, so far so good ... but when I look at the implementation of "forever" i can't wrap my head around :
http://hackage.haskell.org/package/base-4.11.1.0/docs/src/Control.Monad.html...
forever a = let a' = a *> a' in a'
`*>` is the same as `>>` (if you have ever met `>>`), with the difference that the latter works on monads only. In do notation, we could write
forever :: Monad m => m a -> m b forever a = let a' = do a a' in do a'
Which should be easier to get: we `do` a'; a' is nothing but a and a', so we do a and then a', which is nothing but a and a', etc.
Does this explanation feel right?
As soon as you can, familiarise yourself what the do notation desugars to: in my opinion plain operators are clearer -F _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
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Francesco Ariis -
Olivier Revollat