[Bug middle-end/67618] malloc+memset optimization breaks code

["daniel.gutson at tallertechnologies dot com" <gcc-bugzilla at gcc dot gnu dot org>]

https://gcc.gnu.org/bugzilla/show_bug.cgi?id=67618

--- Comment #6 from Daniel Gutson <daniel.gutson at tallertechnologies dot com> ---
(In reply to Andrew Pinski from comment #5)
> (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.

I meant: how do you know what the program will do next? Maybe it will write
memory. See below please.

> 
> > 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.

That's why the 'if (ptr != NULL)' should not be ignored, which currently is.
The 'if' prevents the UB.

> 
> Also calloc is sometimes faster than a malloc/memset due to knowing if the
> kernel will zero out the memory, etc.

This is not under discussion.