[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,

More information about the Haskell-Cafe mailing list