C ++打印chrono :: duration的天，小时，分钟等

问题描述

我有以下代码：

// #includes for <chrono>,<iostream>,etc.

using namespace std;
using namespace chrono;

int main(int argc,char* argv[]) {
    auto sysstart = system_clock::Now();

    LogInit();  // Opens a log file for reading/writing
    // Last shutdown time log entry is marked by a preceding null byte
    logFile.ignore(numeric_limits<streamsize>::max(),'\0');

    if (logFile.fail() || logFile.bad()) {
        // Calls GetLastError() and logs the error code and message
        LogError("main");
    }

    // Parse the timestamp at the start of the shutdown log entry
    tm end = { 0 };
    logFile >> get_time(&end,"[%d-%m-%Y %T");

    if (logFile.fail() || logFile.bad()) {
        // Same as above. Param is name of function within which error occurred
        LogError("main");
    }

    // Finally,we have the last shutdown time as a chrono::time_point
    auto sysEnd = system_clock::from_time_t(mktime(&end));

    // Calculate the time for which the system was inactive
    auto sysInactive = sysstart - sysEnd;
    auto hrs = duration_cast<hours>(sysInactive).count();
    auto mins = duration_cast<minutes>(sysInactive).count() - hrs * 60;
    auto secs = duration_cast<seconds>(sysInactive).count() - (hrs * 3600) - mins * 60;
    auto ms = duration_cast<milliseconds>(sysInactive).count() - (hrs * 3600000)
        - (mins * 60000) - secs * 1000;

    return 0;
}

它可以工作，但是很难看，而且太冗长了。 仅使用STL函数执行此操作的更简单方法？ MTIA：-）

编辑：我意识到，按照@ idclev463035818的注释，完整的代码毕竟不是那么复杂，我已经添加了缺少的内容。

解决方法

当您无法使用fmtlib或等待C ++ 20 <format>时，您至少想尽可能地延迟cout()的调用。另外，让<chrono>为您处理计算。两种措施都可以改善摘要的简洁性：

const auto hrs = duration_cast<hours>(sysInactive);
const auto mins = duration_cast<minutes>(sysInactive - hrs);
const auto secs = duration_cast<seconds>(sysInactive - hrs - mins);
const auto ms = duration_cast<milliseconds>(sysInactive - hrs - secs);

输出：

cout << "System inactive for " << hrs.count() <<
    ":" << mins.count() <<
    ":" << secs.count() <<
    "." << ms.count() << endl;

请注意，您还可以定义实用程序模板，

template <class Rep,std::intmax_t num,std::intmax_t denom>
auto chronoBurst(std::chrono::duration<Rep,std::ratio<num,denom>> d)
{
    const auto hrs = duration_cast<hours>(d);
    const auto mins = duration_cast<minutes>(d - hrs);
    const auto secs = duration_cast<seconds>(d - hrs - mins);
    const auto ms = duration_cast<milliseconds>(d - hrs - secs);

    return std::make_tuple(hrs,mins,secs,ms);
}

结合结构化绑定具有很好的用例：

const auto [hrs,ms] = chronoBurst(sysInactive);