On Fri, Jun 10, 2016 at 3:11 AM, Richard Biener <rguent...@suse.de> wrote:
> On Thu, 9 Jun 2016, Jason Merrill wrote:
>
>> On Thu, Jun 9, 2016 at 3:39 AM, Richard Biener <rguent...@suse.de> wrote:
>> > On Thu, 9 Jun 2016, Jakub Jelinek wrote:
>> >
>> >> On Thu, Jun 09, 2016 at 08:50:15AM +0200, Richard Biener wrote:
>> >> > On Wed, 8 Jun 2016, Jason Merrill wrote:
>> >> >
>> >> > > On Wed, Jun 8, 2016 at 11:16 AM, Marc Glisse <marc.gli...@inria.fr>
>> >> > > wrote:
>> >> > > > On Wed, 8 Jun 2016, Richard Biener wrote:
>> >> > > >
>> >> > > >> The following works around PR70992 but the issue came up repeatedly
>> >> > > >> that we are not very consistent in preserving the undefined
>> >> > > >> behavior
>> >> > > >> of division or modulo by zero. Ok - the only inconsistency is
>> >> > > >> that we fold 0 % x to 0 but not 0 % 0 (with literal zero).
>> >> > > >>
>> >> > > >> After folding is now no longer done early in the C family FEs the
>> >> > > >> number of diagnostic regressions with the patch below is two.
>> >> > > >>
>> >> > > >> FAIL: g++.dg/cpp1y/constexpr-sfinae.C -std=c++14 (test for excess
>> >> > > >> errors)
>> >> > >
>> >> > > Yep. We don't want to fold away undefined behavior in a constexpr
>> >> > > function, since constexpr evaluation wants to detect undefined
>> >> > > behavior and treat the expression as non-constant in that case.
>> >> >
>> >> > Hmm. So 0 / x is not constant because x might be zero but 0 * x is
>> >> > constant because it can never invoke undefined behavior? Does this mean
>> >> > that 0 << n is not const because n might be too large (I suppose
>> >> > 0 << 12000 is not const already)? Is 0 * (-x) const? x might be
>> >> > INT_MIN.
>> >>
>> >> E.g. for the shifts the C++ FE has cxx_eval_check_shift_p which should
>> >> optionally warn and/or set *non_constant_p. 0 * (-INT_MIN) would be
>> >> non-constant too, etc.
>> >> constexpr int foo (int x) { return -x; }
>> >> constexpr int bar (int x) { return 0 * (-x); }
>> >> constexpr int a = foo (-__INT_MAX__ - 1);
>> >> constexpr int b = bar (-__INT_MAX__ - 1);
>> >> shows that we don't diagnose the latter though, most likely because
>> >> constexpr evaluation is done on the folded tree.
>>
>> I don't think there's any folding before constexpr evaluation in this
>> case, since this doesn't involve a call.
>>
>> >> So, either whatever cp_fold does (which uses fold* underneath) should
>> >> avoid
>> >> folding such cases, or the constexpr evaluation should be done on a copy
>> >> made before folding. Then cp_fold doesn't have to prohibit optimizations
>> >> and it is a matter of constexpr.c routines to detect all the undefined
>> >> behavior. After all, I think doing constexpr evaluation on unfolded trees
>> >> will have also the advantage of better diagnostic locations.
>> >
>> > Yes, I think constexpr diagnostic / detection should be done on
>> > unfolded trees (not sure why we'd need to a copy here, just do folding
>> > later?). If cp_fold already avoids folding 0 * (-x) then it should
>> > simply avoid folding 0 / x or 0 % x as well (for the particular issue
>> > this thread started on).
>>
>> The issue is with constexpr functions, which get delayed folding like
>> any other function before they are used in constant expression
>> evaluation. The copy would be to preserve the unfolded trees through
>> cp_fold_function. I've been thinking about doing this anyway; this
>> may be the motivation to go ahead with it.
>
> Or delay the folding until genericization - that is, after all parsing
> is complete. Not sure what this would do to memory usage or compile-time
> when we keep all unfolded bodies (or even re-use them in template
> instantiation).
Genericization happens at the end of each (non-template) function, not
at EOF. Do you mean delay until gimplification?
Jason
