[cfe-dev] [llvm-dev] Should isnan be optimized out in fast-math mode?

Richard Smith via cfe-dev cfe-dev at lists.llvm.org
Fri Sep 17 01:19:22 PDT 2021

On Thu, 16 Sept 2021 at 11:18, Arthur O'Dwyer via cfe-dev <
cfe-dev at lists.llvm.org> wrote:

> On Thu, Sep 16, 2021 at 1:31 PM Serge Pavlov <sepavloff at gmail.com> wrote:
>> On Tue, Sep 14, 2021 at 12:50 AM Serge Pavlov <sepavloff at gmail.com>
>> wrote:
>>> On Mon, Sep 13, 2021 at 11:46 PM Chris Tetreault <ctetreau at quicinc.com>
>>> wrote:
>>>> … is guaranteed to work, and I read that fast-math enables the compiler
>>>> to reason about constructs like `x + 0` being equal to `x`, then I’m going
>>>> to be very confused when:
>>> You are right, this was a bad idea. Compiler may optimize out `isnan`
>>> but only when it deduces that the value cannot be NaN, but not due to the
>>> user's promise. It is especially important for `isinf`. Addition of two
>>> finite values may produce infinity and there is no universal way to predict
>>> it. It is probably not an issue for types like float or double, but ML
>>> cores use halfs or even minifloats, where overflow is much more probable.
>>> If in the code:
>>> ```
>>> float r = a + b;
>>> if (isinf(r)) {...
>>> ```
>>> `isinf` were optimized out just because -ffinite-math-only is in effect,
>>> the user cannot check if overflow did not occur.
>> Rules proposed by Richard are also formulated using arguments, not
>> results. Now there is no intention to optimize such a case.
> Infinity (HUGE_VAL) is already not NaN, so this example doesn't have
> anything to do with the NaN cases being discussed.
> However, let's rephrase as a NaN situation:
>     bool f1(float a, float b) {
>         float r = a + b;
>         return isnan(r);
>     }
>     bool result = f1(-HUGE_VAL, HUGE_VAL);  // expect "true"
> Here, `a + b` can produce quiet-NaN (if `a` is -HUGE_VAL and `b`
> is +HUGE_VAL).
> By Richard Smith's -ffast-math proposal as I understand it, this quiet-NaN
> result would be treated "as if" it were a signaling NaN.
> Under IEEE 754, no operation ever produces a signaling NaN, so
> unfortunately IEEE 754 can't guide us here; but intuitively, I think we'd
> all say that merely *producing* a signaling NaN would not itself *cause*
> a signal. So we store the quiet-NaN result in `r`.
> Then we ask whether `isnan(r)`. The quiet-NaN result in `r` is used. By
> Richard Smith's -ffast-math proposal as I understand it, any operation
> *would* produce an unspecified result if it would raise a signal; but in
> fact `isnan(r)` is a non-signaling operation, so even though we're treating
> quiet-NaN as signaling-NaN, isnan(r) never raises any signal. So this code
> has *well-defined behavior* in -ffast-math mode.
> (And because the code's behavior is well-defined, therefore `isnan(r)` has
> its usual meaning. When `r` holds a quiet-NaN, as in this case, `isnan(r)`
> will correctly return `true`.)

For what it's worth, this is indeed exactly what I meant. Thanks for

> I've googled, but failed to discover, whether comparison against a
> signaling NaN is expected to signal. That is,
>     bool f2(float a, float b) {
>         float r = a + b;
>         return (r != r);
>     }
>     bool result = f2(-HUGE_VAL, HUGE_VAL);  // expect "true" in IEEE754
> mode, but perhaps "false" in -ffast-math mode
> I'm really *hoping* that comparison is a signaling operation. If it is,
> then according to Richard Smith's proposal as I understand it, the compiler
> would be free to optimize `(r != r)` into `(false)` in -ffast-math mode.
> (And, as a corollary, the compiler would *not* generally be free to
> transform `isnan(r)` into `(r != r)`, because the latter expression has
> more preconditions than the former.)
> –Arthur
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