Re: operator new returns nonzero

[Marc Glisse <marc dot glisse at inria dot fr>]

On Sat, 7 Sep 2013, Mike Stump wrote:

> On Sep 7, 2013, at 12:37 PM, Gabriel Dos Reis <gdr@integrable-solutions.net> wrote:
>
> > On Sat, Sep 7, 2013 at 2:27 PM, Marc Glisse <marc.glisse@inria.fr> wrote:
> > > On Sat, 7 Sep 2013, Mike Stump wrote:
> > >
> > > > On Sep 7, 2013, at 3:33 AM, Marc Glisse <marc.glisse@inria.fr> wrote:
> > > > > this patch teaches the compiler that operator new, when it can throw,
> > > > > isn't allowed to return a null pointer.
> > > >
> > > >
> > > > You sure:
> > > >
> > > > @item -fcheck-new
> > > > @opindex fcheck-new
> > > > Check that the pointer returned by @code{operator new} is non-null
> > > > before attempting to modify the storage allocated.  This check is
> > > > normally unnecessary because the C++ standard specifies that
> > > > @code{operator new} only returns @code{0} if it is declared
> > > > @samp{throw()}, in which case the compiler always checks the
> > > > return value even without this option.  In all other cases, when
> > > > @code{operator new} has a non-empty exception specification, memory
> > > > exhaustion is signalled by throwing @code{std::bad_alloc}.  See also
> > > > @samp{new (nothrow)}.
> > > >
> > > > ?
> > >
> > >
> > > Thanks, I didn't know that option. But it doesn't do the same.
> >
> > Indeed.
>
> Can this throw:
>
> void *operator new (long unsigned int s) {
>  return 0;
> }
>
> ?  Is this allowed to return 0?

I think using this function is illegal. It isn't marked noexcept, so it 
isn't allowed to return 0.

--
Marc Glisse