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From: dewar@merv.cs.nyu.edu (Robert Dewar)
Subject: Re: OO, C++, and something much better!
Date: 1997/01/25
Message-ID: <dewar.854208814@merv>#1/1
X-Deja-AN: 212159349
references: <dewar.852652911@merv> <E3rr9G.989@thomsoft.com>
 <dewar.852928982@merv> <5bie2d$eat$1@news.nyu.edu>
organization: New York University
newsgroups: 
 comp.lang.c++,comp.lang.smalltalk,comp.lang.eiffel,comp.lang.ada,comp.object,comp.software-eng
Date: 1997-01-25T00:00:00+00:00
List-Id: <comp.lang.ada>


Richard said, replying to me

>>equally if you do something like
>>
>>   (x = p; x < &p[10]; p += 3)
>>
>>you can find you have an infinite loop, all quite in accordance with the
>>C semantics, although somewhat surprising.
>
>This case must work.  Such a reference to the end of an object is required
>to be treated like a reference within the object.

(note, the p += 3 should of course be x += 3, as I noted in a followup
 post, but I assume Richard understands this typo)

Chuckle, chuckle, it is nice to be able to catch Richard out like this :-)

Richard, you did not look at the example carefully enough. The fact that
you think it works even after I warned that it does not is a perfect
example of how serious this pitfall is!

>   (x = p; x < &p[10]; x += 3)


The problem in my example is that I am bumping x by 3 each time. Sure
it would work if the 3 is replaced by 1, because of the special kludge
in the ANSI standard to allow &p[10] but the increment of 3 causes the
value of x to be further past the end of the array, and this is not
required to work.