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From: diamond@jit345.swstokyo.dec.com (Norman Diamond)
Newsgroups: comp.lang.c++
Subject: Re: the SUN way.. *&$^#%)
Message-ID: <1991Jan2.053835.11860@jituha.enet.dec.com>
Date: 2 Jan 91 05:38:35 GMT
References: <1990Dec20.114821@roadster.bellcore.com> <3069@lupine.NCD.COM> <246@salt.bellcore.com> <3089@lupine.NCD.COM>
Sender: news@jituha.enet.dec.com (USENET News System)
Reply-To: diamond@jit345.enet.dec.com (Norman Diamond)
Organization: Digital Equipment Corporation Japan , Tokyo
Lines: 30

In article <3089@lupine.NCD.COM> rfg@NCD.COM (Ron Guilmette) writes:
+In article <246@salt.bellcore.com> garnett@shera.UUCP (Michael Garnett) writes:
++Q: If is is "always legal to cast any pointer-to-T to a void*" then why
++	won't C++ accept compar(const T*, const T*)? (Assuming that my compiler
++	was correct in telling me that it wouldn't)  Why does C++ allow
++	a call of the form 'free(T*)' when free has a parameter profile of
++	'free(void*)'?  I see no difference in these 2 cases. Both involve
++	accepting T* where a void* is wanted.
+
+You ask "why?"  The answer is "Because those are the rules".  There are
+important differences in the two cases you cite.

There is a real reason, which Mr. Guilmette almost hinted at.
It is exactly true that any pointer-to-T can be CAST (converted) to a void*.
On many machines, the representation is the same, so the cast is a no-op.

However, on some machines, if a pointer-to-T is picked up and used bitwise,
without conversion, as a void*, it might not point to the desired location.
So if you have declarations:
  int a(char f(void*));
  char g(float*));
then you want a diagnostic for this:
  x = a(g);
Otherwise a will pass g something that g can't use.

[Mr. Guilmette:  Please kindly use your "r" or similar command to send me
e-mail.  I wish to see if news is working on this machine.  Thank you.]

Norman Diamond       diamond@tkov50.enet.dec.com
If this were the company's opinion, I wouldn't be allowed to post it.