c++ - C++20 std::chrono::duration 格式的缺点

Question

我需要以std::chrono::duration易于阅读的方式打印可能超过 24 小时的内容，例如"2 d. 09:10:11".

C++20std::chrono::time_point和std::chrono::duration 格式（基于 Howard Hinnant 的日期库）包含大量有用的格式说明符，但在std::chrono::duration. 最接近的是%j，但它用零填充它。一天中的小时数也没有说明符。最接近的是%H（但它也包括全天的小时数）和%OH（但它在断言小时数小于 24 时崩溃）。

我错过了什么吗？

向 C++20 添加更多说明符是否有意义std::format？

到目前为止，我正在使用解决方法：

#include <chrono>
#include <cstring> // std::strncmp
#include <date.h> // use Howard Hinnant's date library until C++20 is supported by compilers
#include <string>

// Workaround for C++20 std::chrono::format's inability to produce number of days in std::chrono::duration.
// Temporary format specifier for this is `%J`. It must be in the beginning of the format string.
template<typename Rep, typename Period>
std::string duration_to_str(const char* format, const std::chrono::duration<Rep, Period>& duration)
{
    if (std::strncmp(format, "%J", 2))
        return date::format(format, duration);
    static const auto one_day = std::chrono::hours(24);
    const size_t days = std::chrono::duration_cast<std::chrono::hours>(duration).count() / 24;
    return std::to_string(days) + date::format(format + 2, duration - days * one_day);
}

测试：

#include <iostream>

template<typename Rep, typename Period>
void test_dur_to_str(const std::chrono::duration<Rep, Period>& duration)
{
    std::cout << duration_to_str("%T", duration) << ":\n"
              << duration_to_str("%j d. %T", duration) << '\n'
              << duration_to_str("%J d. %T", duration) << "\n\n";
}

int main()
{
    test_dur_to_str(std::chrono::seconds(24 * 60 * 60 - 1));
    test_dur_to_str(std::chrono::seconds(24 * 60 * 60));
    test_dur_to_str(std::chrono::seconds(91297));
    test_dur_to_str(std::chrono::seconds(2 * 24 * 60 * 60 - 1));
    test_dur_to_str(std::chrono::seconds(2 * 24 * 60 * 60));
    test_dur_to_str(std::chrono::seconds(2 * 24 * 60 * 60 + 1));
    test_dur_to_str(std::chrono::seconds(13 * 24 * 60 * 60 + 5));
    test_dur_to_str(std::chrono::seconds(365 * 24 * 60 * 60 + 5));
    test_dur_to_str(std::chrono::seconds(366 * 24 * 60 * 60 + 5));
    test_dur_to_str(std::chrono::seconds(367 * 24 * 60 * 60 + 5));
}

输出：

23:59:59:
000 d. 23:59:59
0 d. 23:59:59

24:00:00:
001 d. 24:00:00
1 d. 00:00:00

25:21:37:
001 d. 25:21:37
1 d. 01:21:37

47:59:59:
001 d. 47:59:59
1 d. 23:59:59

48:00:00:
002 d. 48:00:00
2 d. 00:00:00

48:00:01:
002 d. 48:00:01
2 d. 00:00:01

312:00:05:
013 d. 312:00:05
13 d. 00:00:05

8760:00:05:
365 d. 8760:00:05
365 d. 00:00:05

8784:00:05:
366 d. 8784:00:05
366 d. 00:00:05

8808:00:05:
367 d. 8808:00:05
367 d. 00:00:05

标签： c++formattingstdchronoc++20

Answer 1

正如 Howard Hinnant 在他的评论中所建议的那样，我想向那些可能觉得它有用的人展示简化版本的变通方法：

#include <chrono>
#include <cstring> // std::strncmp
#include <date.h> // use Howard Hinnant's date library until C++20 is supported by compilers
#include <string>

template<typename Rep, typename Period>
std::string duration_to_str(const char* format, const std::chrono::duration<Rep, Period>& duration)
{
    if (std::strncmp(format, "%J", 2))
        return date::format(format, duration);
    const auto days = std::chrono::duration_cast<date::days>(duration);
    return std::to_string(days.count()) + date::format(format + 2, duration - days);
}