Path: utzoo!utgpu!jarvis.csri.toronto.edu!mailrus!uunet!aplcen!samsung!think!ames!sun-barr!newstop!sun!brahmand!grover
From: grover%brahmand@Sun.COM (Vinod Grover)
Newsgroups: comp.object
Subject: Re: Continuations
Message-ID: <128709@sun.Eng.Sun.COM>
Date: 2 Dec 89 04:15:36 GMT
References: <2664@bingvaxu.cc.binghamton.edu> <9624@pyr.gatech.EDU> <1623@odin.SGI.COM> <128489@sun.Eng.Sun.COM> <31794@news.Think.COM>
Sender: news@sun.Eng.Sun.COM
Reply-To: grover@sun.UUCP (Vinod Grover)
Organization: Sun Microsystems, Mountain View
Lines: 18

In article <31794@news.Think.COM> barmar@think.com writes:
>In article <128489@sun.Eng.Sun.COM> grover@sun.UUCP (Vinod Grover) writes:
>>I do not know if you are talking about first-class continuations or not, but
>>in languages which are stack-based first-class continuations are not trivial
>>to add. (By stack-based I mean that the procedure entry exit sequance
>>behaves in a stack-like fashion)
>
>This is a circular argument.  It's the very existence of first-class
>continuations in Scheme-like languages that causes them not to be
>stack-based.  If Scheme didn't have continuations then it would be
>stack-based, according to your definition.
One can also argue that it is the existence of lexical scoping, and
higher-order functions that cause a language not to be stack based. In this
sense, continuations are simply functions whose environment may be embedded
in other functions. 

Vinod Grover
Sun Microsystems


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