在不掉帧的情况下录制视频的正确方法是将这两项任务（帧获取和帧序列化）隔离开来，使它们不会相互影响（特别是为了使序列化的波动不会占用捕获帧的时间，这必须无延迟地发生，以防止丢帧）。

这可以通过将序列化（对帧进行编码并将其写入视频文件）委派给单独的线程，并使用某种同步队列将数据馈送到工作线程来实现。

下面是一个简单的例子，展示了如何做到这一点。由于我只有一台相机，而不是你的那种，我会简单地使用网络摄像头并复制帧，但一般原则也适用于你的场景。

示例代码

一开始我们有一些包括：

#include <opencv2/opencv.hpp>
#include <chrono>
#include <condition_variable>
#include <iostream>
#include <mutex>
#include <queue>
#include <thread>
// ============================================================================
using std::chrono::high_resolution_clock;
using std::chrono::duration_cast;
using std::chrono::microseconds;
// ============================================================================

已同步队列

第一步是定义我们的同步队列，我们将使用它与编写视频的工作线程进行通信。

我们需要的主要功能是：

• 将新图像推入队列
• 从队列中弹出图像，在队列为空时等待
• 当我们完成时，可以取消所有挂起的弹出窗口

我们使用std::queue来保存cv::Mat实例，使用std::mutex来提供同步。std::condition_variable用于在图像已插入队列（或取消标志集）时通知消费者，而简单的布尔标志用于通知取消。

最后，我们使用空的struct cancelled作为从pop()抛出的异常，因此我们可以通过取消队列来干净地终止工作进程。

// ============================================================================
class frame_queue
{
public:
    struct cancelled {};
public:
    frame_queue();
    void push(cv::Mat const& image);
    cv::Mat pop();
    void cancel();
private:
    std::queue<cv::Mat> queue_;
    std::mutex mutex_;
    std::condition_variable cond_;
    bool cancelled_;
};
// ----------------------------------------------------------------------------
frame_queue::frame_queue()
    : cancelled_(false)
{
}
// ----------------------------------------------------------------------------
void frame_queue::cancel()
{
    std::unique_lock<std::mutex> mlock(mutex_);
    cancelled_ = true;
    cond_.notify_all();
}
// ----------------------------------------------------------------------------
void frame_queue::push(cv::Mat const& image)
{
    std::unique_lock<std::mutex> mlock(mutex_);
    queue_.push(image);
    cond_.notify_one();
}
// ----------------------------------------------------------------------------
cv::Mat frame_queue::pop()
{
    std::unique_lock<std::mutex> mlock(mutex_);
    while (queue_.empty()) {
        if (cancelled_) {
            throw cancelled();
        }
        cond_.wait(mlock);
        if (cancelled_) {
            throw cancelled();
        }
    }
    cv::Mat image(queue_.front());
    queue_.pop();
    return image;
}
// ============================================================================

存储工人

下一步是定义一个简单的storage_worker，它将负责从同步队列中获取帧，并将它们编码到视频文件中，直到队列被取消。

我添加了简单的计时，所以我们对编码帧花费了多少时间有了一些了解，也对控制台进行了简单的日志记录，因此我们对程序中发生的事情有了一些想法。

// ============================================================================
class storage_worker
{
public:
    storage_worker(frame_queue& queue
        , int32_t id
        , std::string const& file_name
        , int32_t fourcc
        , double fps
        , cv::Size frame_size
        , bool is_color = true);
    void run();
    double total_time_ms() const { return total_time_ / 1000.0; }
private:
    frame_queue& queue_;
    int32_t id_;
    std::string file_name_;
    int32_t fourcc_;
    double fps_;
    cv::Size frame_size_;
    bool is_color_;
    double total_time_;
};
// ----------------------------------------------------------------------------
storage_worker::storage_worker(frame_queue& queue
    , int32_t id
    , std::string const& file_name
    , int32_t fourcc
    , double fps
    , cv::Size frame_size
    , bool is_color)
    : queue_(queue)
    , id_(id)
    , file_name_(file_name)
    , fourcc_(fourcc)
    , fps_(fps)
    , frame_size_(frame_size)
    , is_color_(is_color)
    , total_time_(0.0)
{
}
// ----------------------------------------------------------------------------
void storage_worker::run()
{
    cv::VideoWriter writer(file_name_, fourcc_, fps_, frame_size_, is_color_);
    try {
        int32_t frame_count(0);
        for (;;) {
            cv::Mat image(queue_.pop());
            if (!image.empty()) {
                high_resolution_clock::time_point t1(high_resolution_clock::now());
                ++frame_count;
                writer.write(image);
                high_resolution_clock::time_point t2(high_resolution_clock::now());
                double dt_us(static_cast<double>(duration_cast<microseconds>(t2 - t1).count()));
                total_time_ += dt_us;
                std::cout << "Worker " << id_ << " stored image #" << frame_count
                    << " in " << (dt_us / 1000.0) << " ms" << std::endl;
            }
        }
    } catch (frame_queue::cancelled& /*e*/) {
        // Nothing more to process, we're done
        std::cout << "Queue " << id_ << " cancelled, worker finished." << std::endl;
    }
}
// ============================================================================

正在处理

最后，我们可以把这些放在一起。

我们从初始化和配置视频源开始。然后，我们创建两个frame_queue实例，每个实例对应一个图像流。接下来，我们创建两个storage_worker实例，每个队列一个。为了让事情变得有趣，我为每个设置了不同的编解码器。

下一步是创建并启动工作线程，它将执行每个storage_worker的run()方法。让我们的消费者做好准备，我们可以开始从相机捕捉帧，并将它们提供给frame_queue实例。如上所述，我只有一个源，所以我将同一帧的副本插入到两个队列中。

注意：我需要使用cv::Mat的clone()方法来进行深度复制，否则出于性能原因，我将插入对OpenCV VideoCapture使用的单个缓冲区的引用。这意味着工作线程将获得对该单个映像的引用，并且对该共享映像缓冲区的访问将不同步。您需要确保这种情况不会在您的场景中发生。

一旦我们读取了适当数量的帧（您可以实现您想要的任何其他类型的停止条件），我们就会取消工作队列，等待工作线程完成。

最后，我们写一些关于不同任务所需时间的统计数据。

// ============================================================================
int main()
{
    // The video source -- for me this is a webcam, you use your specific camera API instead
    // I only have one camera, so I will just duplicate the frames to simulate your scenario
    cv::VideoCapture capture(0);
    // Let's make it decent sized, since my camera defaults to 640x480
    capture.set(CV_CAP_PROP_FRAME_WIDTH, 1920);
    capture.set(CV_CAP_PROP_FRAME_HEIGHT, 1080);
    capture.set(CV_CAP_PROP_FPS, 20.0);
    // And fetch the actual values, so we can create our video correctly
    int32_t frame_width(static_cast<int32_t>(capture.get(CV_CAP_PROP_FRAME_WIDTH)));
    int32_t frame_height(static_cast<int32_t>(capture.get(CV_CAP_PROP_FRAME_HEIGHT)));
    double video_fps(std::max(10.0, capture.get(CV_CAP_PROP_FPS))); // Some default in case it's 0
    std::cout << "Capturing images (" << frame_width << "x" << frame_height
        << ") at " << video_fps << " FPS." << std::endl;
    // The synchronized queues, one per video source/storage worker pair
    std::vector<frame_queue> queue(2);
    // Let's create our storage workers -- let's have two, to simulate your scenario
    // and to keep it interesting, have each one write a different format
    std::vector <storage_worker> storage;
    storage.emplace_back(std::ref(queue[0]), 0
        , std::string("foo_0.avi")
        , CV_FOURCC('I', 'Y', 'U', 'V')
        , video_fps
        , cv::Size(frame_width, frame_height)
        , true);
    storage.emplace_back(std::ref(queue[1]), 1
        , std::string("foo_1.avi")
        , CV_FOURCC('D', 'I', 'V', 'X')
        , video_fps
        , cv::Size(frame_width, frame_height)
        , true);
    // And start the worker threads for each storage worker
    std::vector<std::thread> storage_thread;
    for (auto& s : storage) {
        storage_thread.emplace_back(&storage_worker::run, &s);
    }
    // Now the main capture loop
    int32_t const MAX_FRAME_COUNT(10);
    double total_read_time(0.0);
    int32_t frame_count(0);
    for (; frame_count < MAX_FRAME_COUNT; ++frame_count) {
        high_resolution_clock::time_point t1(high_resolution_clock::now());
        // Try to read a frame
        cv::Mat image;
        if (!capture.read(image)) {
            std::cerr << "Failed to capture image.n";
            break;
        }
        // Insert a copy into all queues
        for (auto& q : queue) {
            q.push(image.clone());
        }        
        high_resolution_clock::time_point t2(high_resolution_clock::now());
        double dt_us(static_cast<double>(duration_cast<microseconds>(t2 - t1).count()));
        total_read_time += dt_us;
        std::cout << "Captured image #" << frame_count << " in "
            << (dt_us / 1000.0) << " ms" << std::endl;
    }
    // We're done reading, cancel all the queues
    for (auto& q : queue) {
        q.cancel();
    }
    // And join all the worker threads, waiting for them to finish
    for (auto& st : storage_thread) {
        st.join();
    }
    if (frame_count == 0) {
        std::cerr << "No frames captured.n";
        return -1;
    }
    // Report the timings
    total_read_time /= 1000.0;
    double total_write_time_a(storage[0].total_time_ms());
    double total_write_time_b(storage[1].total_time_ms());
    std::cout << "Completed processing " << frame_count << " images:n"
        << "  average capture time = " << (total_read_time / frame_count) << " msn"
        << "  average write time A = " << (total_write_time_a / frame_count) << " msn"
        << "  average write time B = " << (total_write_time_b / frame_count) << " msn";
    return 0;
}
// ============================================================================

控制台输出

运行这个小示例，我们在控制台中获得以下日志输出，以及磁盘上的两个视频文件。

注意：由于这实际上比捕获快得多，所以我在storage_worker中添加了一些等待，以更好地显示分离。

Capturing images (1920x1080) at 20 FPS.
Captured image #0 in 111.009 ms
Captured image #1 in 67.066 ms
Worker 0 stored image #1 in 94.087 ms
Captured image #2 in 62.059 ms
Worker 1 stored image #1 in 193.186 ms
Captured image #3 in 60.059 ms
Worker 0 stored image #2 in 100.097 ms
Captured image #4 in 78.075 ms
Worker 0 stored image #3 in 87.085 ms
Captured image #5 in 62.061 ms
Worker 0 stored image #4 in 95.092 ms
Worker 1 stored image #2 in 193.187 ms
Captured image #6 in 75.074 ms
Worker 0 stored image #5 in 95.093 ms
Captured image #7 in 63.061 ms
Captured image #8 in 64.061 ms
Worker 0 stored image #6 in 102.098 ms
Worker 1 stored image #3 in 201.195 ms
Captured image #9 in 76.074 ms
Worker 0 stored image #7 in 90.089 ms
Worker 0 stored image #8 in 91.087 ms
Worker 1 stored image #4 in 185.18 ms
Worker 0 stored image #9 in 82.08 ms
Worker 0 stored image #10 in 94.092 ms
Queue 0 cancelled, worker finished.
Worker 1 stored image #5 in 179.174 ms
Worker 1 stored image #6 in 106.102 ms
Worker 1 stored image #7 in 105.104 ms
Worker 1 stored image #8 in 103.101 ms
Worker 1 stored image #9 in 104.102 ms
Worker 1 stored image #10 in 104.1 ms
Queue 1 cancelled, worker finished.
Completed processing 10 images:
  average capture time = 71.8599 ms
  average write time A = 93.09 ms
  average write time B = 147.443 ms
  average write time B = 176.673 ms

可能的改进

目前，在序列化无法跟上相机生成新图像的速度的情况下，无法防止队列过满。为队列大小设置一些上限，并在推送框架之前签入生产者。你需要决定如何处理这种情况。