c++ - c++中的精确采样

Question

我想以每秒 4000 次的频率对从 gpio 获得的值进行采样，目前我正在做类似的事情:

std::vector<int> sample_a_chunk(unsigned int rate, unsigned int block_size_in_seconds) {
    std::vector<std::int> data;
    constexpr unsigned int times = rate * block_size_in_seconds;
    constexpr unsigned int delay = 1000000 / rate; // microseconds
    for (int j=0; j<times; j++) {
      data.emplace_back(/* read the value from the gpio */);
      std::this_thread::sleep_for(std::chrono::microseconds(delay));
    }
    return data;
}

然而根据引用文献，sleep_for 保证等待至少指定的时间量。

我怎样才能让我的系统等待准确的时间，或者至少达到尽可能高的准确性？如何确定系统的时间分辨率？

Answer 1

我认为最好的办法是使用绝对时间以避免漂移。

像这样:

std::vector<int> sample_a_chunk(unsigned int rate,
    unsigned int block_size_in_seconds)
{
    using clock = std::chrono::steady_clock;

    std::vector<int> data;

    const auto times = rate * block_size_in_seconds;
    const auto delay = std::chrono::microseconds{1000000 / rate};

    auto next_sample = clock::now() + delay;

    for(int j = 0; j < times; j++)
    {
        data.emplace_back(/* read the value from the gpio */);

        std::this_thread::sleep_until(next_sample);

        next_sample += delay; // don't refer back to clock, stay absolute
    }
    return data;
}