On Mon, Aug 20, 2018 at 06:27:45PM +0100, Ramsay Jones wrote:
>
>
> On 16/08/18 23:12, Luc Van Oostenryck wrote:
> > In an expression like this, if the inner constant (K) and
> > the mask are disjoint ((K & M) == 0), then the OR with
> > the constant is unneeded and can be optimized away since
> > the constant will be 'cleared' by the mask, giving: OP(x, K)
> >
> > For example, code like:
> > int foo(int x)
> > {
> > return (x | 0xfffff000) & 0xfff;
> > }
>
> Sorry to be dense, but I'm lost again. The above example implies
> to me that OP == &, C == 0xfffff000, K == 0xfff. So what is M?
> Yes ((K & C) == 0) leaving OP(x, K) => (x & 0xfff).
>
> Again where is M coming from?
Yes, sorry, it seems that I changed my mind about using C or K
and ended with things wrong.
In the title OP((x | C), K), 'K' is the real constant of the operation
(or the new instruction size if OP is TRUNC).
In the description, I wrongly used 'K' instead of 'C' and so
((K & M) == 0) should have been ((C & M) == 0).
The whole description here above should be:
In an expression like this, where OP is AND, the effective
mask M is K itself. Then if this mask and the inner constant
(C) are disjoint ((C & M) == 0), then OR instruction is
unneeded and can be optimized away since the constant will
be 'cleared' by the mask, giving: AND(x, M)
Much more consise (and clear!):
if
(C & M) = 0
then
(x | C) & M = (x & M)
Thanks for noticing this.
-- Luc
