C++ 如何以 Java 获取的方式获取自 1970 年以来以毫秒为单位的当前时间戳
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原文地址: http://stackoverflow.com/questions/19555121/
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How to get current timestamp in milliseconds since 1970 just the way Java gets
提问by AKIWEB
In Java, we can use System.currentTimeMillis()to get the current timestamp in Milliseconds since epoch time which is -
在 Java 中，我们可以使用System.currentTimeMillis()自纪元时间以来以毫秒为单位的当前时间戳，即 -
the difference, measured in milliseconds, between the current time and midnight, January 1, 1970 UTC.
当前时间与 UTC 时间 1970 年 1 月 1 日午夜之间的差值（以毫秒为单位）。
In C++ how to get the same thing?
在 C++ 中如何得到同样的东西？
Currently I am using this to get the current timestamp -
目前我正在使用它来获取当前时间戳 -
struct timeval tp;
gettimeofday(&tp, NULL);
long int ms = tp.tv_sec * 1000 + tp.tv_usec / 1000; //get current timestamp in milliseconds
cout << ms << endl;
This looks right or not?
这看起来对还是不对？
回答by Oz.
If you have access to the C++ 11 libraries, check out the std::chronolibrary. You can use it to get the milliseconds since the Unix Epoch like this:
如果您有权访问 C++ 11 库，请查看该std::chrono库。您可以使用它来获取自 Unix Epoch 以来的毫秒数，如下所示：
#include <chrono>
// ...
using namespace std::chrono;
milliseconds ms = duration_cast< milliseconds >(
system_clock::now().time_since_epoch()
);
回答by Trying
回答by kayleeFrye_onDeck
This answer is pretty similar to Oz.'s, using <chrono>for C++ -- I didn't grab it from Oz. though...
这个答案与Oz.'s非常相似，<chrono>用于 C++——我没有从 Oz 那里获取它。尽管...
I picked up the original snippet at the bottom of this page, and slightly modified it to be a complete console app. I love using this lil' ol' thing. It's fantastic if you do a lot of scripting and need a reliable tool in Windows to get the epoch in actual milliseconds without resorting to using VB, or some less modern, less reader-friendly code.
我选择了本页底部的原始代码段，并将其稍作修改，使其成为一个完整的控制台应用程序。我喜欢使用这个小东西。如果您编写了大量脚本并且需要在 Windows 中使用可靠的工具来在实际毫秒内获得纪元，而无需使用 VB 或一些不太现代、不太友好的代码，那就太棒了。
#include <chrono>
#include <iostream>
int main() {
unsigned __int64 now = std::chrono::duration_cast<std::chrono::milliseconds>(std::chrono::system_clock::now().time_since_epoch()).count();
std::cout << now << std::endl;
return 0;
}
回答by rli
If using gettimeofday you have to cast to long long otherwise you will get overflows and thus not the real number of milliseconds since the epoch: long int msint = tp.tv_sec * 1000 + tp.tv_usec / 1000; will give you a number like 767990892 which is round 8 days after the epoch ;-).
如果使用 gettimeofday 你必须转换为 long long 否则你会溢出，因此不是自纪元以来的实际毫秒数：long int msint = tp.tv_sec * 1000 + tp.tv_usec / 1000; 会给你一个像 767990892 这样的数字，这是纪元后 8 天左右；-)。
int main(int argc, char* argv[])
{
struct timeval tp;
gettimeofday(&tp, NULL);
long long mslong = (long long) tp.tv_sec * 1000L + tp.tv_usec / 1000; //get current timestamp in milliseconds
std::cout << mslong << std::endl;
}
回答by Alessandro Pezzato
Since C++11 you can use std::chrono:
从 C++11 开始，您可以使用std::chrono：
- get current system time:
std::chrono::system_clock::now() - get time since epoch:
.time_since_epoch() - translate the underlying unit to milliseconds:
duration_cast<milliseconds>(d) - translate
std::chrono::millisecondsto integer (uint64_tto avoid overflow)
- 获取当前系统时间：
std::chrono::system_clock::now() - 获取自纪元以来的时间：
.time_since_epoch() - 将底层单位转换为毫秒：
duration_cast<milliseconds>(d) - 转换
std::chrono::milliseconds为整数（uint64_t以避免溢出）
#include <chrono>
#include <cstdint>
#include <iostream>
uint64_t timeSinceEpochMillisec() {
using namespace std::chrono;
return duration_cast<milliseconds>(system_clock::now().time_since_epoch()).count();
}
int main() {
std::cout << timeSinceEpochMillisec() << std::endl;
return 0;
}
回答by Owen
Include <ctime>and use the timefunction.
包含<ctime>并使用该time函数。
