无法验证从 time_point 到 tm 和 tm 回到 time_point 的转换

Question

我创建了一个当前 time_point 并将其转换为结构 tm 并打印了它的值。现在将这个 tm 结构转换为 time_point。在比较第一个和第二个 time_point 时，可以看出它们是不同的。但是structure的值是完全一样的

有人能指出我哪里做错了吗？

#include <iostream>
#include <ctime>
#include <chrono>

using namespace std;
using namespace std::chrono;

system_clock::time_point toTimePoint(struct tm tim)
{
    return std::chrono::system_clock::from_time_t(mktime(&tim));
}

tm toTm(system_clock::time_point tp)
{
    time_t tmt = system_clock::to_time_t(tp);
    struct tm * tim = localtime(&tmt);
    struct tm newTim(*tim);
    cout << "Info: " << tim->tm_mday << "/" << tim->tm_mon << "/" << tim->tm_year << " " << tim->tm_hour << ":" << tim->tm_min << ":" << tim->tm_sec << endl;
    cout << "Is Daylight saving: " << tim->tm_isdst << " wday: " << tim->tm_wday << " yday: " << tim->tm_yday << endl;
    return newTim;
}

int _tmain(int argc, _TCHAR* argv[])
{
    system_clock::time_point tp = system_clock::now();

    struct tm tmstruct = toTm(tp);
    system_clock::time_point newtp = toTimePoint(tmstruct);
    cout << "Time comparison: " << (tp == newtp) << endl;

    toTm(newtp);
}

输出：

信息：8/4/115 16:26:20 是夏令时: 0 wday: 5 yday: 127

时间比较：0

信息：8/4/115 16:26:20 是夏令时: 0 wday: 5 yday: 127

Answer 1

四舍五入。 time_point 的分辨率比 time_t 高得多。 time_t 只是秒数，而 time_point 是由系统定义的。例如，在 linux libstdc++ 上，它是纳秒。

例如，您所做的与以下类似

  float f = 4.25;
  int i = (int)f; // i is 4
  std::cout << i << std::endl;
  float f2 = i; // f2 is 4.0 
  std::cout << (f == f2) << std::endl; // false
  int i2 = (int)f2; // i2 is also 4
  std::cout << i2 << std::endl;

无法验证从 time_point 到 tm 和 tm 回到 time_point 的转换

Can't verify the conversion from time_point to tm and tm back to time_point

c++

std

ctime

chrono