3

我正在寻找一种boost::posix_time::time_duration以毫秒为精度输出 a 的简洁解决方案：应该正好有 3 个小数秒数字。默认格式产生 6 个小数位（或者没有，如果它们都是 0）：

#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;
}

使用方面和一些变通方法，我可以获得所需的输出。但是该解决方案仅禁用了我无法根据需要进行配置的日期时间库的部分，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？

4

0 回答 0