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Control characters in # macro.
- To: MOON at MIT-MC, CWH at MIT-MC, GLS at MIT-MC, KMP at MIT-MC, NIL at MIT-MC, BUG-LISP at MIT-MC, ALAN at MIT-AI
- Subject: Control characters in # macro.
- From: Carl W. Hoffman <CWH at MIT-MC>
- Date: Tue, 11 Sep 79 14:16:00 GMT
- Original-date: 11 September 1979 10:16-EDT
Date: 10 SEP 1979 1830-EDT
From: MOON at MIT-MC (David A. Moon)
Date: 10 September 1979 11:08-EDT
From: Carl W. Hoffman <CWH at MIT-MC>
Subject: Control characters in # macro.
To: GLS at MIT-MC, KMP at MIT-MC, NIL at MIT-MC, BUG-LISP at MIT-MC,
Well, this isn't right. One of the motivations for using #/ and #\ is
to be independent of the character set of the particular system being used.
Clearly, #\RUBOUT should read as 207 on the Lisp Machine and in QCMP, and
as 177 in MacLisp. The same should be true of "control" characters. In
MacLisp, though, they could read as ascii characters or in the ITS 12-bit
No. Anything that uses the ascii control characters cannot be independent
of character set, since Lisp machine lisp does not even have them.
If a program needs to use #^ in Maclisp, no amount of syntax will make
it work unchanged on the Lisp machine.
The idea of a character called Control-C is the same when typed on the Lisp
Machine as when typed on an ascii keyboard. I simply want ONE way of
writing this object. There's no reason why a piece of code which does
(= (TYI) #/C) can't work on both the Lisp Machine and in TOPS-20 MacLisp.
The problem which Alan tries to solve with his #^ hack is the fact that two
character sets are available in ITS MacLisp. #/C can read as two different
things -- Control-C as represented in ascii or as represented in the ITS
12-bit character set. This could be controlled by a read-time switch. Anyone
hacking both character sets would presumably canonicalize to a single form.