Question

我正在寻找一个简洁的解决方案来输出精度为毫秒的boost::posix_time::time_duration：应该有正好3个小数 - 秒的数字。默认格式产生6个小数位（如果全部为0则不产生）：

#include <boost/date_time.hpp>
#include <iostream>

int main()
{
    // Define some duration in milliseconds:
    int64_t start_msecs((((40 * 60) + 3) * 60 + 2) * 1000 + 1);

    // The same as time_duration:
    boost::posix_time::time_duration start_time =
        boost::posix_time::milliseconds(start_msecs);

    // No suitable format (for MP4Box chapter starts): ////////////////////
    std::cout << "Wrong format: "
        << std::setprecision(3) // <-- No effect!?
        << start_time << std::endl;
    // Output: "Wrong format: 40:03:02.001000"
    // Required format      : 40:03:02.001

    return 0;
}

使用facet和一些解决方法，我可以得到所需的输出。但该解决方案仅禁用日期时间库的部分，我可以根据自己的需要进行配置，并用低级实现替换它们：

#include <boost/date_time.hpp>
#include <iostream>

int main()
{
    // Define some duration in milliseconds:
    int64_t start_msecs((((40 * 60) + 3) * 60 + 2) * 1000 + 1);

    // The same as time_duration:
    boost::posix_time::time_duration start_time =
            boost::posix_time::milliseconds(start_msecs);

    // Define output format without fractional seconds:
    boost::posix_time::time_facet *output_facet =
        new boost::posix_time::time_facet();
    output_facet->time_duration_format("%O:%M:%S");

    // Imbue cout with format for duration output:
    std::cout.imbue(std::locale(std::locale::classic(), output_facet));

    // Only the milliseconds:
    int64_t msecs_only = start_msecs % 1000;

    // Render duration with exactly 3 fractional-second digits: ///////////
    std::cout << "Working: "
        << start_time << "."
        << std::setw(3) << std::right << std::setfill('0')
        << msecs_only << std::endl;
    // Output: "Working: 40:03:02.001"

    return 0;
}

实现所需输出的推荐方法是什么？

如何以毫秒精度格式化输出的boost :: date_time持续时间？

0 个答案: