On Mon, 25 Apr 2022, David Laight wrote:
> From: Mikulas Patocka
> > Sent: 25 April 2022 12:04
> >
> > On Mon, 25 Apr 2022, David Laight wrote:
> >
> > > From: Linus Torvalds
> > > > Sent: 24 April 2022 22:42
> > > >
> > > > On Sun, Apr 24, 2022 at 2:37 PM Linus Torvalds
> > > > <torvalds@xxxxxxxxxxxxxxxxxxxx> wrote:
> > > > >
> > > > > Finally, for the same reason - please don't use ">> 8". Because I do
> > > > > not believe that bit 8 is well-defined in your arithmetic. The *sign*
> > > > > bit will be, but I'm not convinced bit 8 is.
> > > >
> > > > Hmm.. I think it's ok. It can indeed overflow in 'char' and change the
> > > > sign in bit #7, but I suspect bit #8 is always fine.
> > > >
> > > > Still, If you want to just extend the sign bit, ">> 31" _is_ the
> > > > obvious thing to use (yeah, yeah, properly "sizeof(int)*8-1" or
> > > > whatever, you get my drift).
> > >
> > > Except that right shifts of signed values are UB.
> > > In particular it has always been valid to do an unsigned
> > > shift right on a 2's compliment negative number.
> > >
> > > David
> >
> > Yes. All the standard versions (C89, C99, C11, C2X) say that right shift
> > of a negative value is implementation-defined.
> >
> > So, we should cast it to "unsigned" before shifting it.
>
> Except that the intent appears to be to replicate the sign bit.
>
> If it is 'implementation defined' (rather than suddenly being UB)
The standard says "If E1 has a signed type and a negative value, the
resulting value is implementation-defined."
So, it's not undefined behavior.
> it might be that the linux kernel requires sign propagating
> right shifts of negative values.
It may be that some code in the Linux kernel already assumes that right
shifts keep the sign. It's hard to say if such code exists.
BTW. ubsan warns about left shift of negative values, but it doesn't warn
about right shift of negative values.
> This is typically what happens on 2's compliment systems.
> But not all small cpu have the required shift instruction.
> OTOH all the ones bit enough to run Linux probably do.
> (And gcc doesn't support '1's compliment' or 'sign overpunch' cpus.)
>
> The problem is that the compiler writers seem to be entering
> a mindset where they are optimising code based on UB behaviour.
> So given:
> void foo(int x)
> {
> if (x >> 1 < 0)
> return;
> do_something();
> }
> they decide the test is UB, so can always be assumed to be true
> and thus do_something() is compiled away.
>
> David
If it's implementation-defined (rather than undefined), the compiler
shouldn't do such optimization.
The linux kernel uses "-fno-strict-overflow" which disables some of these
UB optimizations.
Mikulas
