C++ 如何从 <chrono> 获取持续时间，如 int 毫秒和浮点秒数？
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How to get duration, as int milli's and float seconds from <chrono>?
提问by EddieV223
I'm trying to use chrono library for timers and durations.
我正在尝试将 chrono 库用于计时器和持续时间。
I want to be able to have a Duration frameStart;( from app start )
and a Duration frameDelta;( time between frames )
我希望能够有一个Duration frameStart;（从应用程序开始）和一个Duration frameDelta;（帧之间的时间）
I need to be able to get the frameDeltaduration as milliseconds and float seconds.
我需要能够以frameDelta毫秒和浮动秒为单位获得持续时间。
How do you do this with the new c++11 <chrono>libraries? I've been working on it and googling ( information is sparse ). The code is heavily templated and requires special casts and things, I can't figure out how to use this library correctly.
你如何使用新的 c++11<chrono>库来做到这一点？我一直在研究它并使用谷歌搜索（信息很少）。代码是大量模板化的，需要特殊的演员表和东西，我不知道如何正确使用这个库。
回答by Howard Hinnant
Is this what you're looking for?
这是你要找的吗？
#include <chrono>
#include <iostream>
int main()
{
typedef std::chrono::high_resolution_clock Time;
typedef std::chrono::milliseconds ms;
typedef std::chrono::duration<float> fsec;
auto t0 = Time::now();
auto t1 = Time::now();
fsec fs = t1 - t0;
ms d = std::chrono::duration_cast<ms>(fs);
std::cout << fs.count() << "s\n";
std::cout << d.count() << "ms\n";
}
which for me prints out:
这对我来说打印出来：
6.5e-08s
0ms
回答by Jonathan Wakely
I don't know what "milliseconds and float seconds" means, but this should give you an idea:
我不知道“毫秒和浮点秒”是什么意思，但这应该给你一个想法：
#include <chrono>
#include <thread>
#include <iostream>
int main()
{
auto then = std::chrono::system_clock::now();
std::this_thread::sleep_for(std::chrono::seconds(1));
auto now = std::chrono::system_clock::now();
auto dur = now - then;
typedef std::chrono::duration<float> float_seconds;
auto secs = std::chrono::duration_cast<float_seconds>(dur);
std::cout << secs.count() << '\n';
}
回答by Billy The Kid
Taking a guess at what it is you're asking for. I'm assuming by millisecond frame timer you're looking for something that acts like the following,
猜一猜你要的是什么。我假设通过毫秒帧计时器您正在寻找类似于以下内容的东西，
double mticks()
{
struct timeval tv;
gettimeofday(&tv, 0);
return (double) tv.tv_usec / 1000 + tv.tv_sec * 1000;
}
but uses std::chronoinstead,
而是使用std::chrono，
double mticks()
{
typedef std::chrono::high_resolution_clock clock;
typedef std::chrono::duration<float, std::milli> duration;
static clock::time_point start = clock::now();
duration elapsed = clock::now() - start;
return elapsed.count();
}
Hope this helps.
希望这可以帮助。
回答by Chris Drew
In AAA styleusing the explicitly typed initializer idiom:
#include <chrono>
#include <iostream>
int main(){
auto start = std::chrono::high_resolution_clock::now();
// Code to time here...
auto end = std::chrono::high_resolution_clock::now();
auto dur = end - start;
auto i_millis = std::chrono::duration_cast<std::chrono::milliseconds>(dur);
auto f_secs = std::chrono::duration_cast<std::chrono::duration<float>>(dur);
std::cout << i_millis.count() << '\n';
std::cout << f_secs.count() << '\n';
}
