On Mon, Aug 20, 2018 at 06:44:49PM +0100, Ramsay Jones wrote:
>
>
> On 16/08/18 23:12, Luc Van Oostenryck wrote:
> > If the effective mask (C & M) is different than the outer mask C,
> > then this later can be replaced by the smaller (C & M), giving:
> > OP((x | M'), K) with M' == (C & M).
> >
> > For example, code like:
> > int foo(int x)
> > {
> > return (x | 0xfffffff0) & 0xfff;
> > }
>
> C = 0xfffffff0, OP = &, K = 0xfff, (C & K) = 0xff0, so
> OP((x | M'), K) where M' = (C & K).
Indeed but I see it more as:
* OP is AND/&, so M = K
* C = 0xfffffff0, M = 0xfff, M' = (C & M) = 0xff0
and the simplification is:
((x | C) & M) -> ((x | M') & M)
It should be noted that a better simplification here should be:
((x | C) & M) -> (x & M')
but:
* this is not generaly true, it would thus need a new guard
* it can only be done if the OP is AND (while the whole series
is about unified handling of AND/TRUNC/LSR/SHL).
* at first I thought this kinf of simplification would not be
worth (no intructions removed, just remove the inner mask by
a smaller one) but on some code I used it was much more effective
than I thought (and some others that I thought would be quite
effective were not), it all depend on the exact code, of course.
-- Luc
