用C ++ 11打印優雅時間

Question

這就是我想在C ++ 11中做的事情：給定兩個時間點（例如計時類）作為std::chrono::steady_clock::now() ，優雅地打印它們的時差，例如：

1 day 4 hours 3 minutes 45 seconds

要么

32 minutes 54 seconds 345 milliseconds

請注意，我對使用put_time不感興趣，因為我想從最重要的單位時間開始打印。 我知道，它只是打印差異的解決方案，但它並不漂亮：我正在尋找一個優雅的解決方案:)

干杯!

Answer 1

持續時間可以算術。

#include <chrono>
#include <iostream>
#include <thread>

int main(){
    using namespace std::chrono;
    using day_t = duration<long, std::ratio<3600 * 24>>;
    auto start = system_clock::now();
    std::this_thread::sleep_for(seconds(1));
    auto end = system_clock::now();
    auto dur = end - start;
    auto d = duration_cast<day_t>(dur);
    auto h = duration_cast<hours>(dur -= d);
    auto m = duration_cast<minutes>(dur -= h);
    auto s = duration_cast<seconds>(dur -= m);
    auto ms = duration_cast<seconds>(dur -= s);
    std::cout << d.count() << " days, "
        << h.count() << " hours, "
        << m.count() << " minutes, "
        << s.count() << " seconds, "
        << ms.count() << " milliseconds\n";

    return 0;
}

輸出：

在此輸入圖像描述

可能的重復：從C ++中的std :: chrono :: time_point中提取年/月/日等

Answer 2

template<typename T>
void print_time_diff(std::ostream& out, T prior, T latter)
{
    namespace sc = std::chrono;
    auto diff = sc::duration_cast<sc::milliseconds>(latter - prior).count();
    auto const msecs = diff % 1000;
    diff /= 1000;
    auto const secs = diff % 60;
    diff /= 60;
    auto const mins = diff % 60;
    diff /= 60;
    auto const hours = diff % 24;
    diff /= 24;
    auto const days = diff;

    bool printed_earlier = false;
    if (days >= 1) {
        printed_earlier = true;
        out << days << (1 != days ? " days" : " day") << ' ';
    }
    if (printed_earlier || hours >= 1) {
        printed_earlier = true;
        out << hours << (1 != hours ? " hours" : " hour") << ' ';
    }
    if (printed_earlier || mins >= 1) {
        printed_earlier = true;
        out << mins << (1 != mins ? " minutes" : " minute") << ' ';
    }
    if (printed_earlier || secs >= 1) {
        printed_earlier = true;
        out << secs << (1 != secs ? " seconds" : " second") << ' ';
    }
    if (printed_earlier || msecs >= 1) {
        printed_earlier = true;
        out << msecs << (1 != msecs ? " milliseconds" : " millisecond");
    }
}

http://ideone.com/bBNHQp

Answer 3

這是一個使用可變參數模板和遞歸的易於擴展的解決方案。 它定義了ostream& operator<<(ostream&, const duration&)以方便使用。

#include <chrono>
#include <iostream>
#include <tuple>

using day_t = std::chrono::duration<long long, std::ratio<3600 * 24>>;

template<typename> struct duration_traits {};

#define DURATION_TRAITS(Duration, Singular, Plural) \
template<> struct duration_traits<Duration> { \
    constexpr static const char* singular = Singular; \
    constexpr static const char* plural = Plural; \
}

DURATION_TRAITS(std::chrono::milliseconds, "millisecond", "milliseconds");
DURATION_TRAITS(std::chrono::seconds, "second", "seconds");
DURATION_TRAITS(std::chrono::minutes, "minute", "minutes");
DURATION_TRAITS(std::chrono::hours, "hour", "hours");
DURATION_TRAITS(day_t, "day", "days");

using divisions = std::tuple<std::chrono::milliseconds, 
                             std::chrono::seconds, 
                             std::chrono::minutes, 
                             std::chrono::hours, 
                             day_t>;

namespace detail {
template<typename...> struct print_duration_impl_ {};

template<typename Head, typename... Tail>
struct print_duration_impl_<Head, Tail...> {
    template <typename Duration>
    static bool print(std::ostream& os, Duration& dur) {
        const auto started_printing = print_duration_impl_<Tail...>::print(os, dur);

        const auto n = std::chrono::duration_cast<Head>(dur);
        const auto count = n.count();

        if (count == 0) {
            return started_printing;
        }

        if (started_printing) {
            os << ' ';
        }

        using traits = duration_traits<Head>;
        os << count << ' ' << (count == 1 ? traits::singular : traits::plural);
        dur -= n;

        return true;
    }
};

template<>
struct print_duration_impl_<> {
    template <typename Duration>
    static bool print(std::ostream& os, Duration& dur) {
        return false;
    }
};

template<typename...> struct print_duration {};

template<typename... Args>
struct print_duration<std::tuple<Args...>> {
    template<typename Duration>
    static void print(std::ostream& os, Duration dur) {
        print_duration_impl_<Args...>::print(os, dur);
    }
};
}

template<typename Rep, typename Period>
std::ostream& operator<<(std::ostream& os, const std::chrono::duration<Rep, Period>& dur) {
    detail::print_duration<divisions>::print(os, dur);
    return os;
}

通過專門化duration_traits並將類型插入分區中的正確位置來添加新的持續時間。 例如，添加10毫秒的jiffy類型將涉及：

using jiffy_t = std::chrono::duration<long long, std::centi>;
DURATION_TRAITS(jiffy_t, "jiffy", "jiffies");

using divisions = std::tuple<std::chrono::milliseconds, 
                             jiffy_t,
                             std::chrono::seconds, 
                             std::chrono::minutes, 
                             std::chrono::hours, 
                             day_t>;

三行代碼還不錯！

在ideone.com上的實例。

用C ++ 11打印優雅時間

問題描述

3 個解決方案

解決方案1
4 2014-02-27 12:49:23

解決方案2
3 2014-02-27 09:37:21

解決方案3
3 2014-02-27 17:31:44

用C ++ 11打印優雅時間

問題描述

3 個解決方案

解決方案1 4 2014-02-27 12:49:23

解決方案2 3 2014-02-27 09:37:21

解決方案3 3 2014-02-27 17:31:44

解決方案1
4 2014-02-27 12:49:23

解決方案2
3 2014-02-27 09:37:21

解決方案3
3 2014-02-27 17:31:44