[Haskell-cafe] Re: Can we come out of a monad?

Edward Z. Yang ezyang at MIT.EDU
Tue Aug 10 02:40:38 EDT 2010


Excerpts from Ertugrul Soeylemez's message of Tue Aug 10 02:31:14 -0400 2010:
> There is no evalCont, there is runCont:
> 
>   runCont :: (a -> r) -> Cont r a -> r
> 
> Note that Cont/ContT computations result in a value of type 'r':
> 
>   newtype Cont r a = Cont ((a -> r) -> r)

Yes, but if you pass in 'id' as the continuation to 'runCont',
the entire expression will result in 'a'.  The continuation monad
doesn't act globally...

Still confused,
Edward


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