Re: *p++ = *q++ undefined? why?

CBFalconer wrote:

Eric Sosman wrote:
CBFalconer wrote:
pete wrote:

... snip ...

As long as (*p) is defined and (*q) is defined,
there is nothing wrong with (*p++ = *q++).

Assuming p != q.

Even if p == q. (Confession: I very nearly made
the same mistake.)

Alright, modify the assumption to:

&p !- &q

&p != &q

OK

--
pete
.

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