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From: jeff@aiai.ed.ac.uk (Jeff Dalton)
Subject: Re: Interpreters (Re: Comparing productivity: LisP against C++ (was Re: Reference Counting))
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References: <RFB.95Jan20150836@cfdevx1.lehman.com> <MATT.95Jan21135219@physics7.berkeley.edu> <XJAM.95Jan21202837@fir.CS.Berkeley.EDU>
Date: Tue, 24 Jan 1995 17:13:35 GMT
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Xref: glinda.oz.cs.cmu.edu comp.lang.c++:109084 comp.lang.lisp:16481

In article <XJAM.95Jan21202837@fir.CS.Berkeley.EDU> xjam@fir.CS.Berkeley.EDU (The Crossjammer) writes:
>>>>>> "Matt" == Matt Austern <matt@physics7.berkeley.edu> writes:
>In article <MATT.95Jan21135219@physics7.berkeley.edu> matt@physics7.berkeley.edu (Matt Austern) writes:
>    Matt> That's not really true.  Every dialect of lisp I've ever heard of
>    Matt> includes the function eval.  Once you have eval, you have a lisp
>    Matt> interpreter; the top-level read-eval-print loop is just a frill.
>
>If you count Scheme as a dialect of Lisp then there's one without eval.

You should count Scheme as a dialect of Lisp (except that it's
misleading to call any Lisp a dialect of Lisp).  If you look at Scheme's
properties as a language, it's not very different from things that 
virtually no one questions are varieties of Lisp (e.g. EuLisp level 0).
This is, moreover, a fairly mainstream view.  For instance, the 
_Revised^4 Report on the Algorithmic Language Scheme_ says Scheme
is a dialect of Lisp.

>However, it does have load which is just bad.

EuLisp doesn't have load or eval.  Neither (so far as I can tell)
does ISLisp (the ISO WG-16 Lisp).

>Dylan has dispensed with that, so there is a Lisp like language that does
>not have to have an embeded interpreter.

Scheme does not have to have an embedded interpreter.  Even Common
Lisp does not have to have an embedded interpreter.

-- jd



