如何正确使用libvlc（C++）从h264 videostream获取帧？

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我有兴趣获得解决方案，我使用libvlc库从h264流获取视频帧。我已回调至libvlc_video_set_format_callbacks和我的回调函数收到以下信息（format_callback）参数：色度：“J420”，宽度：1088，高度：1922如何正确使用libvlc（C++）从h264 videostream获取帧？

当我打电话主

Player p; 
p.play("rtsp://path/to/camera?videocodec=h264");

它打印出错以下错误

chroma "J420" width: 1088 , height: 1922 
[00007fddfc001268] core vout display error: Failed to change zoom 
[00007fddfc001268] core vout display error: Failed to set on top 
[00007fddfc001268] core vout display error: Failed to change source AR 
[h264 @ 0x7fde1c06cea0] error while decoding MB 24 111, bytestream -15 
[swscaler @ 0x7fddfc002ca0] bad dst image pointers 
[swscaler @ 0x7fddfc002ca0] bad dst image pointers

我的猜测是缓冲区及其大小有问题。在哪里，用哪种类型和多大的缓冲区来逐帧获取视频？后来我打算将帧数据转发给QImage。下面是Player.h文件：

const int ResoHeight = 1922; 
const int ResoWidth = 1088; 
const int BytesPerPixel = 3; // not sure about this 

struct ImageData 
{ 
    QVector<unsigned char> raw; 
    QVector<unsigned char> picture; 

    ImageData() 
    { 
    raw.resize(BytesPerPixel * ResoHeight * ResoWidth); 
    picture.resize(BytesPerPixel * ResoHeight * ResoWidth); 
    } 
}; 

class Player : public QObject 
{ 
    Q_OBJECT 
public: 
    explicit Player(QObject *parent = nullptr); 
    ~Player(); 

    void play(const std::string& path); 

signals: 
    void newImage(const QImage& image); 

private: 
    libvlc_instance_t* vlcInstance; 
    libvlc_media_player_t* player; 
    libvlc_media_t* media; 
    ImageData buffer; 
};

Player.cpp是以下几点：

namespace { 
void* lock_frame(void *opaque, void **planes) 
{ 
    ImageData* buf = (ImageData*)(opaque); 
    *planes = buf->raw.data(); 
    return buf->picture.data(); 
} 

void unlock_frame(void *opaque, void *picture, void *const *planes) 
{ 
    // will be logic to announce new image 
} 

unsigned format_callback(void** opaque, char* chroma, unsigned *width, unsigned *height, unsigned *pitches, unsigned *lines) 
{ 
    qDebug() << "chroma:" << QString(chroma) << "width:" << *width << ", height:" << *height; 

    *pitches= (*width) * BytesPerPixel; 
    *lines= *height; 

    return 1; 
} 

} // namespace 

Player::Player(QObject* parent) 
    : QObject(parent) 
    , vlcInstance(libvlc_new(0, nullptr)) 
    , player(libvlc_media_player_new(vlcInstance)) 
    , media(nullptr) 
{ 
} 

Player::~Player() 
{ 
    libvlc_media_player_stop(player); 
    libvlc_media_player_release(player); 
    libvlc_release(vlcInstance); 
} 

void Player::play(const std::string& path) 
{ 
    media = libvlc_media_new_location(vlcInstance, path.c_str()); 
    libvlc_media_player_set_media(player, media); 
    libvlc_media_release(media); 

    libvlc_video_set_callbacks(player, lock_frame, unlock_frame, nullptr, &buffer); 
    libvlc_video_set_format_callbacks(player, format_callback, nullptr); 

    libvlc_media_player_play(player); 
}

来源

2017-06-13 psalong

'buffer'未初始化。你应该在format_callback函数中初始化它。 – seleciii44

'buffer'是ImageData的类型，它在构造函数中被启动。当然，在'format_callback'函数中初始化它是正确的，但它不应该是问题的原因。通过将缓冲区大小调整方法移动到'format_callback'来试验它。同样的错误。 – psalong

对不起，没有意识到。 J420色度是一种平面YUV格式https://wiki.videolan.org/YUV/。这意味着你必须在lock_frame中提供3个图像指针。如果你只是想要一个RGB（RV24）图像看到这个问题https://stackoverflow.com/q/40365353/1632887 – seleciii44

J420色度是planar YUV格式。这意味着您必须在lock_frame函数中提供3维pitches和lines中的format_callback和3个不同的planes指针（针对每个平面）。如果您只需要RGB（RV24）图像，请参阅this question。

来源

2017-06-14 11:27:32 seleciii44

如何正确使用libvlc（C++）从h264 videostream获取帧？

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