Re: [PATCH] locking: type cleanup when accessing fast_read_ctr

[Oleg Nesterov]

On 05/23, Nicholas Mc Guire wrote:
>
> On Wed, 20 May 2015, Oleg Nesterov wrote:
>
> > On 05/19, Nicholas Mc Guire wrote:
> > >
> > > I assumed it would not matter but did not see a simple way of getting it
> > > type clean with unsigned either mainly due to the atomic_t being int and
> > > val in update_fast_ctr() being passed as -1.
> >
> > Perhaps clear_fast_ctr() should have a comment to explain why it returns
> > "int"... even if "unsigned" should work too.
> >
> Might not be into c99 standard far enough but from reviewing 6.5/J.2
> I do not see this argument here.
>
> The "well defined modulo 2**n" behavior for unsigned int can be
> found stated in a few places - but not in the c99 standard for
> arithmetic overflow.
>
> The "well defined overflow behavior" as far as I understand c99,
> *only* applies to left shift operations when overflowing - see 6.5.7 "
> Bitwise shift operators" -> Semantics -> 4) -  further for the counter
> problem this well defined behavior is of little help as the sum would
> be wrong in both cases.
>
> I still do not see the point in the implicit/automatic type conversion
> here and why that should be an advantage - could somone point me to
> the right c99 clauses ?

Sorry, I don't really understand your question...

Once again. Signed overflow is undefined behaviour according to C standard.
That is why ->fast_read_ctr is "unsigned long", this counter can actually
over/underflow if down/up happens on different CPU's.

clear_fast_ctr() returns "signed int" just because this looks better to me,
it can actually return the negative number. If you make it return "unsigned"
you will simply shift this implicit/automatic type conversion to atomic_add()
which accepts "int i".

Let me also quote Linus:

	Now, I doubt you'll find an architecture or C compiler where this will
	actually ever make a difference,

Yes. So this all is actually cosmetic.

Oleg.

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