On July 20, 2017 4:20:00 PM GMT+02:00, Marek Polacek <pola...@redhat.com> wrote:
>On Thu, Jul 20, 2017 at 12:55:10PM +0200, Richard Biener wrote:
>> On Wed, Jul 19, 2017 at 3:55 PM, Marek Polacek <pola...@redhat.com>
>wrote:
>> > On Wed, Jul 19, 2017 at 12:45:12PM +0200, Richard Biener wrote:
>> >> On Tue, Jul 18, 2017 at 6:05 PM, Marek Polacek
><pola...@redhat.com> wrote:
>> >> > We ended up in infinite recursion between extract_muldiv_1 and
>> >> > fold_plusminus_mult_expr, because one turns this expression into
>the other
>> >> > and the other does the reverse:
>> >> >
>> >> > ((2147483648 / 0) * 2) + 2 <-> 2 * (2147483648 / 0 + 1)
>> >> >
>> >> > I tried (unsuccessfully) to fix it in either extract_muldiv_1 or
>> >> > fold_plusminus_mult_expr, but in the end I went with just
>turning (x / 0) + A
>> >> > to x / 0 (and similarly for %), because with that undefined
>division we can do
>> >> > anything and this fixes the issue. Any better ideas?
>> >>
>> >> Heh - I looked at this at least twice as well with no conclusive
>fix...
>> >>
>> >> My final thought was to fold division/modulo by zero to
>__builtin_trap () but
>> >> I didn't get to implement that. I'm not sure if we need to
>preserve
>> >> the behavior
>> >> of raising SIGFPE as I think at least the C standard makes it
>undefined.
>> >> OTOH other languages with non-call-exceptions might want to catch
>division
>> >> by zero.
>> >
>> > It's definitely undefined in C, so there we can do anything we see
>fit, but not
>> > sure about the rest
>> >
>> >> Did you see why the oscillation doesn't happen for
>> >>
>> >> ((2147483648 / A) * 2) + 2 <-> 2 * (2147483648 / A + 1)
>> >>
>> >> ? What's special for the zero constant as divisor?
>> >
>> > I think it comes down to how split_tree splits the expression. For
>the above
>> > we never call associate_trees, i.e., this condition is never true:
>> >
>> > 9647 if (ok
>> > 9648 && (2 < ((var0 != 0) + (var1 != 0)
>> > 9649 + (con0 != 0) + (con1 != 0)
>> > 9650 + (lit0 != 0) + (lit1 != 0)
>> > 9651 + (minus_lit0 != 0) + (minus_lit1 !=
>0))))
>> >
>> > because var0 = so, lit1 = 2, and the rest is null.
>> >
>> > We also don't go into infinite recursion with x / 0 instead of
>2147483648 / 0,
>> > because split_tree will put "x / 0 + 1" into var0, whereas it will
>put
>> > 2147483648 / 0 into con1, because it's TREE_CONSTANT - and so we
>have more than
>> > 2 exprs that are non-null and we end up looping.
>>
>> Ah yeah - now I remeber. Stupid associate relying on TREE_CONSTANT
>...
>>
>> Maybe sth like
>>
>> Index: gcc/fold-const.c
>> ===================================================================
>> --- gcc/fold-const.c (revision 250379)
>> +++ gcc/fold-const.c (working copy)
>> @@ -816,9 +816,9 @@ split_tree (location_t loc, tree in, tre
>> || TREE_CODE (op1) == FIXED_CST)
>> *litp = op1, neg_litp_p = neg1_p, op1 = 0;
>>
>> - if (op0 != 0 && TREE_CONSTANT (op0))
>> + if (op0 != 0 && TREE_CONSTANT (op0) && ! tree_could_trap_p
>(op0))
>> *conp = op0, op0 = 0;
>> - else if (op1 != 0 && TREE_CONSTANT (op1))
>> + else if (op1 != 0 && TREE_CONSTANT (op1) && !
>tree_could_trap_p (op1))
>> *conp = op1, neg_conp_p = neg1_p, op1 = 0;
>>
>> /* If we haven't dealt with either operand, this is not a case
>we can
>> @@ -846,7 +846,7 @@ split_tree (location_t loc, tree in, tre
>> var = negate_expr (var);
>> }
>> }
>> - else if (TREE_CONSTANT (in))
>> + else if (TREE_CONSTANT (in) && ! tree_could_trap_p (in))
>> *conp = in;
>> else if (TREE_CODE (in) == BIT_NOT_EXPR
>> && code == PLUS_EXPR)
>>
>> would help that?
>
>That works for one testcase, but not for another, e.g. this one:
>
>unsigned int *od;
>int
>fn (void)
>{
> return (0 % 0 + 1) * *od * 2; /* { dg-warning "division by zero" } */
>}
>
>because tree_could_trap_p says "no" for "0 % 0 + 1" -- it just sees the
>outer
>PLUS_MINUS and never checks the subexpression with %. Should we change
>that?
>Guess we'd have to call tree_could_trap_p recursively...
No, we shouldn't. Maybe trapping ops shouldn't be marked TREE_CONSTANT?
(Make sure to test with Ada...)
Richard.
> Marek
