如何用C++限制循环中的FPS？

Question

我正在尝试使用带有计时和线程的C++来限制执行交叉检查的循环中的每秒帧数.

这是我的代码:

std::chrono::system_clock::time_point now = std::chrono::system_clock::now();
std::chrono::system_clock::time_point lastFrame = std::chrono::system_clock::now();

while (true)
{
    // Maintain designated frequency of 5 Hz (200 ms per frame)
    now = std::chrono::system_clock::now();
    std::chrono::duration<double, std::milli> delta = now - lastFrame;
    lastFrame = now;

    if (delta.count() < 200.0)
    {
        std::chrono::duration<double, std::milli> delta_ms(200.0 - delta.count());
        auto delta_ms_duration = std::chrono::duration_cast<std::chrono::milliseconds>(delta_ms);
        std::this_thread::sleep_for(std::chrono::milliseconds(delta_ms_duration.count()));
    }

    printf("Time: %f \n", delta.count());

    // Perform intersection test

}

我遇到的问题是delta的每个其他输出显示的是微不足道的数量,而不是我想要的~200 ms /帧:

Time: 199.253200
Time: 2.067700
Time: 199.420400
Time: 2.408100
Time: 199.494200
Time: 2.306200
Time: 199.586800
Time: 2.253400
Time: 199.864000
Time: 2.156500
Time: 199.293800
Time: 2.075500
Time: 201.787500
Time: 4.426600
Time: 197.304100
Time: 4.530500
Time: 198.457200
Time: 3.482000
Time: 198.365300
Time: 3.415400
Time: 198.467400
Time: 3.595000
Time: 199.730100
Time: 3.373400

有关为什么会发生这种情况的任何想法？

Answer 1

如果您考虑代码的工作原理,您会发现它的编写方式与您的代码完全一致.Delta由于代码中的逻辑错误而振荡.

这是发生的事情:

• 我们一开始delta == 0.
• 因为delta小于200,所以你编码sleeps 200 - delta(0) == 200ms.
• 现在,三角洲本身变得接近200(因为你已经测量了睡眠时间和实际工作)并且你睡了200 - delta(200) == 0ms.
• 之后循环重复.

要解决此问题,您无需测量睡眠时间.

这是如何做到的:

#include <iostream>
#include <cstdio>
#include <chrono>
#include <thread>

std::chrono::system_clock::time_point a = std::chrono::system_clock::now();
std::chrono::system_clock::time_point b = std::chrono::system_clock::now();

int main()
{
    while (true)
    {
        // Maintain designated frequency of 5 Hz (200 ms per frame)
        a = std::chrono::system_clock::now();
        std::chrono::duration<double, std::milli> work_time = a - b;

        if (work_time.count() < 200.0)
        {
            std::chrono::duration<double, std::milli> delta_ms(200.0 - work_time.count());
            auto delta_ms_duration = std::chrono::duration_cast<std::chrono::milliseconds>(delta_ms);
            std::this_thread::sleep_for(std::chrono::milliseconds(delta_ms_duration.count()));
        }

        b = std::chrono::system_clock::now();
        std::chrono::duration<double, std::milli> sleep_time = b - a;

        // Your code here

        printf("Time: %f \n", (work_time + sleep_time).count());
    }
}

这段代码给了我一系列稳定的增量:

Time: 199.057206 
Time: 199.053581 
Time: 199.064718 
Time: 199.053515 
Time: 199.053307 
Time: 199.053415 
Time: 199.053164 
Time: 199.053511 
Time: 199.053280 
Time: 199.053283    

Answer 2

这很像Galik的答案,但它保留了OP问题的语法,并没有下载到C API.此外,它为帧持续时间创建了一个自定义单位,我认为这对于可读性非常重要:

#include <chrono>
#include <cstdint>
#include <iostream>
#include <thread>

int
main()
{
    using namespace std;
    using namespace std::chrono;

    using frames = duration<int64_t, ratio<1, 5>>;  // 5Hz
    auto nextFrame = system_clock::now();
    auto lastFrame = nextFrame - frames{1};;

    while (true)
    {
        // Perform intersection test

        this_thread::sleep_until(nextFrame);
        cout << "Time: "  // just for monitoring purposes
             << duration_cast<milliseconds>(system_clock::now() - lastFrame).count()
             << "ms\n";
        lastFrame = nextFrame;
        nextFrame += frames{1};
    }
}

这为我输出:

Time: 200ms
Time: 205ms
Time: 205ms
Time: 203ms
Time: 205ms
Time: 205ms
Time: 200ms
Time: 200ms
Time: 200ms
...

要注意的关键事项:

• 记录5Hz的简明方法: using frames = duration<int64_t, ratio<1, 5>>;
• 使用sleep_until而不是sleep_for,它可以处理完成实际工作所需的时间.
• .count()除了I/O之外没有使用,这里有一个库可以摆脱它.
• 没有手动转换单位(例如/ 1000).
• 没有浮点单位,不是说有什么问题.
• 最少需要指定或依赖显式单位.

通过添加持续时间I/O库,以下是上述代码的更改方式:

#include "chrono_io.h"
#include <chrono>
#include <cstdint>
#include <iostream>
#include <thread>

int
main()
{
    using namespace date;
    using namespace std;
    using namespace std::chrono;

    using frames = duration<int64_t, ratio<1, 5>>;  // 5Hz
    auto nextFrame = system_clock::now();
    auto lastFrame = nextFrame - frames{1};;

    while (true)
    {
        // Perform intersection test

        this_thread::sleep_until(nextFrame);
        // just for monitoring purposes
        cout << "Time: " << system_clock::now() - lastFrame << '\n';
        lastFrame = nextFrame;
        nextFrame += frames{1};
    }
}

输出将根据平台而不同(取决于"原生持续时间" system_clock).在我的平台上,它看起来像这样:

Time: 200042µs
Time: 205105µs
Time: 205107µs
Time: 200044µs
Time: 205105µs
Time: 200120µs
Time: 204307µs
Time: 205136µs
Time: 201978µs
...

Answer 3

我通常做这样的事情:

#include <chrono>
#include <iostream>

int main()
{
    using clock = std::chrono::steady_clock;

    auto next_frame = clock::now();

    while(true)
    {
        next_frame += std::chrono::milliseconds(1000 / 5); // 5Hz

        // do stuff
        std::cout << std::time(0) << '\n'; // 5 for each second

        // wait for end of frame
        std::this_thread::sleep_until(next_frame);
    }
}

输出:(每秒五个值)

1470173964
1470173964
1470173964
1470173964
1470173964
1470173965
1470173965
1470173965
1470173965
1470173965
1470173966
1470173966
1470173966
1470173966
1470173966