Re: no continuations

2003-12-31 Thread ajb
G'day all. Quoting Tomasz Zielonka <[EMAIL PROTECTED]>: > OK. I think I may be getting it now. The point is that MonadCont takes > care of passing the continuation, so you don't have to do it by hand. Is > that right? Precisely. > Happy New Year, And to you and yours. Cheers, Andrew Bromage _

Re: no continuations

2003-12-31 Thread Tomasz Zielonka
On Tue, Dec 30, 2003 at 10:31:57PM -0500, [EMAIL PROTECTED] wrote: > G'day all. > > Quoting Tomasz Zielonka <[EMAIL PROTECTED]>: > > > BTW, the factorial example on > > http://www.haskell.org/hawiki/MonadicContinuationPassingStyle > > seems rather pointless to me, because it doesn't use any met

Re: no continuations

2003-12-30 Thread ajb
G'day all. Quoting Tomasz Zielonka <[EMAIL PROTECTED]>: > BTW, the factorial example on > http://www.haskell.org/hawiki/MonadicContinuationPassingStyle > seems rather pointless to me, because it doesn't use any methods > of MonadCont (like callCC). The only point of the factorial example is to

Re: no continuations

2003-12-30 Thread Amr A Sabry
Kevin S. Millikin <[EMAIL PROTECTED]> wrote: > Oh, sure. I didn't mean to quibble with the idea that continuations > are computational effects. Just wanted to point out that (I think) you > can't macro express mutation with call/cc, unless you've already got > mutation anyway. That's right:

Re: no continuations

2003-12-30 Thread Derek Elkins
On Tue, 30 Dec 2003 10:38:33 -0800 (PST) Ben Rudiak-Gould <[EMAIL PROTECTED]> wrote: > On Tue, 30 Dec 2003, Scott wrote: > > Why does Haskell have no continuations? > > (http://www.haskell.org/hawiki/CoMonad) > > If continuations are incompatible with non-strict semant

RE: no continuations

2003-12-30 Thread Kevin S. Millikin
On Tuesday, December 30, 2003 3:10 PM, Ben Rudiak-Gould [SMTP:[EMAIL PROTECTED] wrote: > Interesting. > > This still violates referential transparency, though. (c 'get) returns > a value or errors out depending on whether (c 'set) has been called yet. Oh, sure. I didn't mean to quibble with th

RE: no continuations

2003-12-30 Thread Ben Rudiak-Gould
On Tue, 30 Dec 2003, Kevin S. Millikin wrote: > On Tuesday, December 30, 2003 12:39 PM, Ben Rudiak-Gould wrote: > > With letrec and unrestricted call/cc you can implement ML-style refs: > > With an *implementation of letrec that uses mutation* and unrestricted > call/cc, you can implement ML-styl

RE: no continuations

2003-12-30 Thread Kevin S. Millikin
On Tuesday, December 30, 2003 12:39 PM, Ben Rudiak-Gould [SMTP:[EMAIL PROTECTED] wrote: > > With letrec and unrestricted call/cc you can implement ML-style refs: With an *implementation of letrec that uses mutation* and unrestricted call/cc, you can implement ML-style ref cells: Petite Chez Sch

Re: no continuations

2003-12-30 Thread Ben Rudiak-Gould
On Tue, 30 Dec 2003, Scott wrote: > Why does Haskell have no continuations? > (http://www.haskell.org/hawiki/CoMonad) > If continuations are incompatible with non-strict semantics, I'd > appreciate an explanation. With letrec and unrestricted call/cc you can implement ML-style

Re: no continuations

2003-12-30 Thread Tomasz Zielonka
On Tue, Dec 30, 2003 at 07:21:08AM -0600, Scott wrote: > Why does Haskell have no continuations? > (http://www.haskell.org/hawiki/CoMonad) See http://www.haskell.org/hawiki/MonadCont BTW, the factorial example on http://www.haskell.org/hawiki/MonadicContinuationPassingStyle seems

Re: no continuations

2003-12-30 Thread Lennart Augustsson
"cached" result from the previous call. It's not an insurmountable problem, but it's pretty hairy. -- Lennart Scott wrote: Why does Haskell have no continuations? (http://www.haskell.org/hawiki/CoMonad) If continuations are incompatible with non-str

no continuations

2003-12-30 Thread Scott
Why does Haskell have no continuations? (http://www.haskell.org/hawiki/CoMonad) If continuations are incompatible with non-strict semantics, I'd appreciate an explanation. ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/ma