[Haskell-cafe] How to make callCC more dynamic

oleg at okmij.org oleg at okmij.org
Sat Aug 27 08:50:36 CEST 2011

> ContT {runContT :: forall r1 . (forall r2 . a-> m r2) -> m r1}
> callCC can be defined, however, you can not run it, and reset couldn't
> type check

Indeed you cannot. As the articles

explain, the answer of _undelimited_ continuation is not available
to the program itself. You really cannot write runUndelimitedCont --
just as you cannot write runIO. Once you in the monad of undelimited
continuations, you cannot get out of it -- just you cannot get out of
IO. Since reset is the composition of runCont and return, reset is not
expressible either.

The article above explains that in detail (see the CPS2 attempt). The
article also shows how to cheat.

This exercise points out that undelimited continuations are really not
useful. In fact, I don't know of any practical application of them.
I'm deeply puzzled why people insist on using callCC given how useless
it is by itself (without other effects such as mutation). If one uses
callCC and runCont, one deals with _delimited_ continuation. Why not
to use shift then, which has a bit more convenient interface.

Do you have a specific code that you want to write using ContT? It
is generally more productive to discuss a concrete example.

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