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Re: can s-exprs represent a graph without eval?
I've never seen this, it's certainly interesting, but how to you detect
that you're just going around in circles?
(Let's say someone passes you v and you aren't suspecting v has this
Geoffrey S. Knauth | http://knauth.org/gsk
On Thursday, Jun 19, 2003, at 13:58 US/Eastern, Shriram Krishnamurthi
> Many extensions to Scheme support this, and I imagine they probably
> obtained their notation from (Common?) Lisp. In Chez Scheme, PLT
> Scheme and others, you can write programs such as this:
>> (define v '#0=(a . #0#))
> so that
>> (car v)
>> (cadr v)
>> (caddr v)
>> (cdr v)
> #0=(a . #0#)
> In general, #n= binds while #n# refers.