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*To*: scheme@mc.lcs.mit.edu*Subject*: Re^2: Scheme Digest #8, Efficiency of Y*From*: zodiac!DUCHAMPS.ADS.COM!rar@ames.arpa (Bob Riemenschneider)*Date*: Thu ,17 Nov 88 17:08:31 EDT*Sender*: scheme-request@mc.lcs.mit.edu

Max Hailpern writes: => There is another reason, in Scheme, not to use Y: it breaks the fact => that you can use procedures as objects. While the R^3R says that a => procedure created by a given lambda at a given time will be eqv to => itself (and implies eq as well, though on looking I see that isn't in => there -- is this a mistake?), the same does not necessarily hold for => the various incarnations of a procedure that Y will churn out (or => rather, that Y is entitled to churn out: presumably in Rozas's => implementation there is indeed only one procedure). => => Perhaps I'm wrong to mix together such disparate worlds as Y and => eqvness of procedures belong to, but I do consider this to be => something of an issue. Does anyone else? Could you provide a specific example? I don't see how Y makes the situation any worse than lambda. Just as you might decide to write (let ((p (lambda (n) ...n...))) ===p===p===) rather than ===(lambda(n) ...n...)===(lambda (n) ...n...)=== because Scheme doesn't guarantee that even syntactically identical lambda terms will test equivalent---Does anyone know why 'operational equivalence' for procedures was defined extensionally, making it uncomputable, rather than intensionally (say, in terms of alpha-convertability)?--, you might decide to write (let ((p (Y (lambda (q) ...q...)))) ===p===p===) rather than ===(Y (lambda(q) ...q...))===(Y (lambda (q) ...q...))=== Arguments against Y that apply equally well to lambda don't count! -- rar

**Follow-Ups**:**Re: Re^2: Scheme Digest #8, Efficiency of Y***From:*titan!dorai@rice.edu (Dorai Sitaram)

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