On Wed, Aug 31, 2005 at 12:09:40AM +0200, Bastian Blank wrote: > On Tue, Aug 30, 2005 at 12:34:01AM -0700, Steve Langasek wrote: > > When passing pointers to 4-byte types to memcpy(), gcc-4.0 generates > > wrong code which assumes that these pointers are aligned at 4-byte > > boundaries for purposes of optimization, ignoring the implicit cast to > > (char *) in the prototype of memcpy().
> There is no implicit cast to char *. memcpy gets two void pointers. Hmm, yes... Unfortunately, it appears that using (void *) is not sufficient to stop gcc-4.0 from peeking at the pointer in this case -- whereas, if memcpy() is implemented as a function, it certainly would be... On Wed, Aug 31, 2005 at 12:05:35AM +0200, Bastian Blank wrote: > On Tue, Aug 30, 2005 at 02:45:41PM -0700, Steve Langasek wrote: > > By any chance, can you provide a reference to the C spec that shows > > gcc's current behavior is valid? Given that traceroute is among the > > programs that breaks under gcc-4.0, it seems to me that the assumption > > that it's safe to use memcpy this way has been around for quite some > > time. > 6.3.2.3, paragraph 7. That paragraph appears to read: A pointer to an object or incomplete type may be converted to a pointer to a different object or incomplete type. If the resulting pointer is not correctly aligned for the pointed-to type, the behavior is undefined. Otherwise, when converted back again, the result shall compare equal to the original pointer. When a pointer to an object is converted to a pointer to a character type, the result points to the lowest addressed byte of the object. So by making the initial cast from char * to the pointer type that requires greater alignment, the programmer is invoking undefined behavior, which means gcc's implementation of memcpy() is allowed by the spec. That's what I wanted to know, thanks. -- Steve Langasek Give me a lever long enough and a Free OS Debian Developer to set it on, and I can move the world. [EMAIL PROTECTED] http://www.debian.org/
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