On Fri, Jul 24, 2015 at 9:48 AM, Kai Tietz <ktiet...@googlemail.com> wrote: > 2015-07-24 7:54 GMT+02:00 Jeff Law <l...@redhat.com>: >> On 07/23/2015 10:33 AM, Segher Boessenkool wrote: >>> >>> On Thu, Jul 23, 2015 at 10:09:49AM -0600, Jeff Law wrote: >>>> >>>> It seems to me in these kind of cases that selection of the canonical >>>> form should be driven by factors outside of which is better for a >>>> particular target. ie, which is simpler >>> >>> >>> I agree. But neither form is simpler here, and we need to have both >>> forms in other contexts, so there is no real benefit to canonicalising. >> >> >> >> a << N ==/!= 0 >> >> Looks like two operations. A shift and a comparison against zero regardless >> of whether or not N is constant. >> >> >> a&(-1>>N) ==/!= 0 >> >> For a varying N, this has a shift, logical and and comparison against zero. >> >> For a constant N obviously the shift collapses to a constant and we're left >> with two operations again. >> >> So for gimple, I'd prefer to see us using the a << N form. >> >> If we need both forms in other contexts, we ought to be looking to eliminate >> that need :-) >> >> If we go to the RTL level, then it's more complex -- it might depend on the >> constant produced by the -1 >> N operation, whether or not the target can >> shift by more than one bit at a time (H8/300 series is limited here for >> example), whether or not one operation sets condition codes in a useful way, >> potentially allowing the comparison to be removed, etc etc. rtx_costs, even >> with its limitations is probably the way to drive selection of form for the >> RTL optimizers. >> >> >> Jeff > > I fully agree. But there are case there is not necessarily a 'better' > representation. For example for sinking converts into shift-operation > can be tricky. > > for 'typea a;' > > (typeb) (a << N) -> ((typeb) a) << N > > 'bitsizeof (typeb) <= N' result is always zero. > 'bitsizeof (typeb) > N' result needs to be calculated. > > But it isn't necessarily easy to say if form '(typeb) (a << N)', or > '((typeb) a) << N' form is to be prefered. It strongly depends on > expression this pattern is used in. > > For RTL this pattern has another issue, as we need to take natural > mode into account here too, > > We should define for such forms our 'normal' representation.
The usual reason for canonicalization is CSE. In this case you'd need to have X = a & (-1 >> N) == 0; Y = a << N == 0; and want to CSE Y. Might be not a very common pattern though. Richard. > Kai
