优雅的C++11时间打印
优雅的C++11时间打印
提问于 2014-02-27 09:18:33
这就是我在C++11中想要做的:给定两个时间点(例如，一个定时类)作为std::chrono::steady_clock::now()，优雅地打印它们的时间差，例如：
1 day 4 hours 3 minutes 45 seconds或
32 minutes 54 seconds 345 milliseconds请注意，我对使用put_time并不感兴趣，因为我想从最重要的时间单位开始打印。我知道，这是一个解决方案，只是打印差异，但这并不漂亮:我正在寻找一个优雅的解决方案:)
干杯!
回答 2
Stack Overflow用户
发布于 2014-02-27 17:31:44
这里有一个易于扩展的解决方案，使用各种模板和递归。为了便于使用，它定义了ostream& operator<<(ostream&, const duration&)。
#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;
}新的持续时间是通过专门化的duration_traits并在分区中正确的位置插入类型来添加的。例如，添加一个10毫秒的吉菲类型将涉及：
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>;三行代码还不错！
在ideone.com上的实例。
Stack Overflow用户
发布于 2014-02-27 09:37:21
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");
}
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/22063979
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