https://gcc.gnu.org/bugzilla/show_bug.cgi?id=67618
Andrew Pinski <pinskia at gcc dot gnu.org> changed:
What |Removed |Added
----------------------------------------------------------------------------
Component|c |middle-end
--- Comment #5 from Andrew Pinski <pinskia at gcc dot gnu.org> ---
(In reply to Daniel Gutson from comment #4)
> (In reply to Andreas Schwab from comment #3)
> > Trying to read the (uninitialized) contents of the allocated memory for x <=
> > 12 would be undefined behaviour, so calling calloc instead does not change
> > the defined behaviour.
>
> Why do you assert that the program will read the memory?
It does not. It just optimizes the code.
> The optimization ignores the 'if' statement (just comment that line and
> see), so if malloc() returns NULL, memset is called unconditionally and will
> try to write to address NULL. The "x > 12" was just an example that this is
> arbitrary. Replace it with something else. The 'if' shall not be ignored.
yes that just undefined behavior when invoking memset with a NULL value, so
replacing it is still fine.
Also calloc is sometimes faster than a malloc/memset due to knowing if the
kernel will zero out the memory, etc.
