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List:       gcc-patches
Subject:    Re: [PATCH] Simplify floating point comparisons
From:       Marc Glisse <marc.glisse () inria ! fr>
Date:       2018-04-30 18:49:59
Message-ID: alpine.DEB.2.21.1804302040170.2839 () stedding ! saclay ! inria ! fr
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On Fri, 12 Jan 2018, Wilco Dijkstra wrote:

> Hi,
>
> Here is the updated version:
>
> This patch implements some of the optimizations discussed in
> https://gcc.gnu.org/bugzilla/show_bug.cgi?id=71026.
>
> Simplify (C / x >= 0.0) into x >= 0.0 with -fno-signed-zeros
> and -ffinite-math-only.  If C is negative the comparison is reversed.
> Only handle >= and <= for now since C / x can underflow if C is small.
>
>
> Simplify (x * C1) > C2 into x > (C2 / C1) with -funsafe-math-optimizations.
> If C1 is negative the comparison is reversed.
>
> OK for commit?
>
> ChangeLog
> 2018-01-10  Wilco Dijkstra  <wdijkstr@arm.com>
> 	    Jackson Woodruff  <jackson.woodruff@arm.com>
>
>    gcc/
> 	PR 71026/tree-optimization
> 	* match.pd: Simplify floating point comparisons.
>
>    gcc/testsuite/
> 	PR 71026/tree-optimization
> 	* gcc.dg/div-cmp-1.c: New test.
> 	* gcc.dg/div-cmp-2.c: New test.
> --
>
> diff --git a/gcc/match.pd b/gcc/match.pd
> index 435125a317275527661fba011a9d26e507d293a6..8a6fee906de6a750201362119862f8326868f26b 100644
> --- a/gcc/match.pd
> +++ b/gcc/match.pd
> @@ -376,6 +376,21 @@ DEFINE_INT_AND_FLOAT_ROUND_FN (RINT)
>  (rdiv @0 (negate @1))
>  (rdiv (negate @0) @1))
>
> +/* Simplify (C / x op 0.0) to x op 0.0 for C != 0, C != Inf/Nan.
> +   Only handle >= and <= since C / x may underflow to zero.  */
> +(for op (le ge)
> +     res_op (lt ge)
> +     neg_op (ge lt)
> + (simplify
> +  (op (rdiv REAL_CST@0 @1) real_zerop@2)
> +  (if (!HONOR_SIGNED_ZEROS (@1) && !HONOR_INFINITIES (@1))
> +   (switch
> +    (if (real_less (&dconst0, TREE_REAL_CST_PTR (@0)))
> +     (res_op @1 @2))
> +    /* For C < 0, use the inverted operator.  */
> +    (if (real_less (TREE_REAL_CST_PTR (@0), &dconst0))
> +     (neg_op @1 @2))))))

Let's try with C = DBL_MIN and x = ąDBL_MAX. I don't believe it involves 
signed zeros or infinities, just an underflow. First, the result depends 
on the rounding mode. And in the default round-to-nearest, both divisions 
give 0, and thus compare the same with 0, but we replace that with a sign 
test on x, where they clearly give opposite answers.

What would be the proper flag to test to check if we care about underflow?

-- 
Marc Glisse
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