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- Newsgroups: comp.lang.ada,comp.lang.c,comp.lang.c++,comp.edu
- Path: news.mathworks.com!gatech!psinntp!psinntp!psinntp!psinntp!twratl!tynor
- From: tynor@atlanta.twr.com (Steve Tynor)
- Subject: Re: ANSI C and POSIX (was Re: C/C++ knocks the crap out of Ada)
- In-Reply-To: c2a192@ugrad.cs.ubc.ca's message of 8 Apr 1996 21:36:36 -0700
- Message-ID: <TYNOR.96Apr9105114@twratl.atlanta.twr.com>
- Sender: tynor@atlanta.twr.com (Steve Tynor)
- Organization: Tower Technology
- References: <JSA.96Feb16135027@organon.com> <4kb2j8$an0@solutions.solon.com>
- <4kbrt5$k3h@mulga.cs.mu.OZ.AU> <dewar.829012486@schonberg>
- <4kcpgkINNcku@keats.ugrad.cs.ubc.ca>
- Date: Tue, 9 Apr 1996 15:51:14 GMT
-
- In article <4kcpgkINNcku@keats.ugrad.cs.ubc.ca> c2a192@ugrad.cs.ubc.ca (Kazimir Kylheku) writes:
-
- | 1.3: If I write the code
- |
- | int i, j;
- |
- | can I assume that (&i + 1) == &j?
- |
- | A: Only sometimes. It's not portable, because in EBCDIC, i and j are
- | not adjacent.
-
- What does this have to do with EBCDIC? i and j in this examples are
- just identifier names of variables -- &i is the address of the
- variable i. I has nothing whatsoever to do with the EBCDIC (or ASCII
- or whatever character set you choose) value of the characters 'i' and
- 'j'. The answer is still "only sometimes", but not for the reason
- given :-). Or did I miss a Date: header of "1 Apr"?
-
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- Programming needn't be like penance or oat bran.
-
- Steve Tynor Internet: Steve.Tynor@atlanta.twr.com
- Tower Technology WWW: http://www.twr.com/
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