std::chrono::round(std::chrono::duration)

 

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std::chrono::duration

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duration::duration duration::operator= duration::count duration::zero duration::min duration::max duration::operator+duration::operator-	duration::operator++duration::operator-- duration::operator+=duration::operator-=duration::operator*=duration::operator/=duration::operator%=
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operator+operator-operator*operator/operator% operator==operator!=operator<operator<=operator>operator>=operator<=> (C++20 前)(C++20) operator<< (C++20)	duration_cast floor (C++17) ceil (C++17) round (C++17) abs (C++17) operator""h (C++14) operator""min (C++14) operator""s (C++14) operator""ms (C++14) operator""us (C++14) operator""ns (C++14) from_stream (C++20)
辅助类
common_type treat_as_floating_point      duration_values	formatter<std::chrono::duration> (C++20) hash<std::chrono::duration> (C++26)

 

在标头 <chrono> 定义
template< class ToDuration, class Rep, class Period > constexpr ToDuration round( const std::chrono::duration<Rep, Period>& d );		(C++17 起)

返回 ToDuration 可以表示的最接近 d 的 {{tt|t} 。若有两个这样的值，则返回偶数值（即使得 t % 2 == 0 的值 t）。

除非 ToDuration 是 std::chrono::duration 的特化且 std::chrono::treat_as_floating_point_v<typename ToDuration::rep> 为 false，否则函数不参与重载决议。

参数

d

-

要转换的时长

返回值

取整到 ToDuration 类型的最接近 d 时长，两值正中的情况取到偶数。

可能的实现

namespace detail { template<class> inline constexpr bool is_duration_v = false; template<class Rep, class Period> inline constexpr bool is_duration_v< std::chrono::duration<Rep, Period>> = true; }   template<class To, class Rep, class Period, class = std::enable_if_t<detail::is_duration_v<To> && !std::chrono::treat_as_floating_point_v<typename To::rep>>> constexpr To round(const std::chrono::duration<Rep, Period>& d) { To t0 = std::chrono::floor<To>(d); To t1 = t0 + To{1}; auto diff0 = d - t0; auto diff1 = t1 - d; if (diff0 == diff1) { if (t0.count() & 1) return t1; return t0; } else if (diff0 < diff1) return t0; return t1; }

示例

运行此代码

#include <iostream>
#include <iomanip>
#include <chrono>
 
int main()
{
    using namespace std::chrono_literals;
    using Sec = std::chrono::seconds;
    for (std::cout << "Duration\tFloor\tRound\tCeil\n"
                      "(ms)\t\t(sec)\t(sec)\t(sec)\n";
        auto const d: {
            +4999ms, +5000ms, +5001ms, +5499ms, +5500ms, +5999ms,
            -4999ms, -5000ms, -5001ms, -5499ms, -5500ms, -5999ms, }) {
        std::cout << std::showpos << d.count() << "\t\t"
                  << std::chrono::floor<Sec>(d).count() << '\t'
                  << std::chrono::round<Sec>(d).count() << '\t'
                  << std::chrono::ceil <Sec>(d).count() << '\n';
    }
}

输出：

Duration   Floor   Round   Ceil
(ms)       (sec)   (sec)   (sec)
+4999      +4      +5      +5
+5000      +5      +5      +5
+5001      +5      +5      +6
+5499      +5      +5      +6
+5500      +5      +6      +6
+5999      +5      +6      +6
-4999      -5      -5      -4
-5000      -5      -5      -5
-5001      -6      -5      -5
-5499      -6      -5      -5
-5500      -6      -6      -5
-5999      -6      -6      -5

参阅

duration_cast (C++11)	转换时长到另一个拥有不同计次间隔的时长 (函数模板)
floor(std::chrono::duration) (C++17)	以向下取整的方式，将一个时长转换为另一个时长 (函数模板)
ceil(std::chrono::duration) (C++17)	以向上取整的方式，将一个时长转换为另一个时长 (函数模板)
round(std::chrono::time_point) (C++17)	转换 time_point 到另一个，就近取整，偶数优先 (函数模板)
roundroundfroundllroundlroundflroundlllroundllroundfllroundl (C++11)(C++11)(C++11)(C++11)(C++11)(C++11)(C++11)(C++11)(C++11)	最接近整数，中间情况下向远离零舍入 (函数)