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<chrono>: optimize chrono::steady_clock::now() #2086

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Merged
Stephan T. Lavavej (StephanTLavavej) merged 4 commits into from
Aug 17, 2021
Merged

<chrono>: optimize chrono::steady_clock::now() #2086

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51bf525
optimize chrono:steady_clock::now()
fsb4000 Jul 30, 2021
87a8d69
use `static` for `_Freq`
fsb4000 Jul 30, 2021
8e1daf4
Revert "use `static` for `_Freq`"
fsb4000 Jul 30, 2021
5bed616
Addressing PR reviews
fsb4000 Jul 30, 2021
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26 changes: 18 additions & 8 deletions stl/inc/chrono
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Original file line number Diff line number Diff line change
Expand Up @@ -705,14 +705,24 @@ namespace chrono {
const long long _Freq = _Query_perf_frequency(); // doesn't change after system boot
const long long _Ctr = _Query_perf_counter();
static_assert(period::num == 1, "This assumes period::num == 1.");
// Instead of just having "(_Ctr * period::den) / _Freq",
// the algorithm below prevents overflow when _Ctr is sufficiently large.
// It assumes that _Freq * period::den does not overflow, which is currently true for nano period.
// It is not realistic for _Ctr to accumulate to large values from zero with this assumption,
// but the initial value of _Ctr could be large.
const long long _Whole = (_Ctr / _Freq) * period::den;
const long long _Part = (_Ctr % _Freq) * period::den / _Freq;
return time_point(duration(_Whole + _Part));
// 10 MHz is a very common QPC frequency on modern PCs. Optimizing for
// this specific frequency can double the performance of this function by
// avoiding the expensive frequency conversion path.
constexpr long long _TenMHz = 10'000'000;
if (_Freq == _TenMHz) {
static_assert(period::den % _TenMHz == 0, "It should never fail.");
constexpr long long _Multiplier = period::den / _TenMHz;
return time_point(duration(_Ctr * _Multiplier));
} else {
// Instead of just having "(_Ctr * period::den) / _Freq",
// the algorithm below prevents overflow when _Ctr is sufficiently large.
// It assumes that _Freq * period::den does not overflow, which is currently true for nano period.
// It is not realistic for _Ctr to accumulate to large values from zero with this assumption,
// but the initial value of _Ctr could be large.
const long long _Whole = (_Ctr / _Freq) * period::den;
const long long _Part = (_Ctr % _Freq) * period::den / _Freq;
return time_point(duration(_Whole + _Part));
}
}
};

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