从多播流中计算 RTP 序列号

Question

我尝试从多播流中计算 RTP 序列号并确定丢失的包。

请看下面的C代码:

int sock, bytesRead;
char buffer[MAXBUFSIZE];
uint16_t seq = 0;
uint16_t eseq = 0;
struct sockaddr_in saddr;
socklen_t socklen;

    for(;;) {
        bytesRead = recvfrom(sock, buffer, MAXBUFSIZE, 0,
                             (struct sockaddr *)&saddr, &socklen);
        if (bytesRead) {
            seq = (buffer[2] << 8)+buffer[3];
            eseq++;
            if (seq != eseq) {
                std::cerr << " SEQ = " << seq << " ESEC = " << eseq << "\n";
                eseq = seq;
            }
        }
    }

我收到了以下结果:

12:47:44 SEQ = 58112 ESEC = 57856
12:47:45 SEQ = 57984 ESEC = 58240
12:47:45 SEQ = 58368 ESEC = 58112
12:47:45 SEQ = 58240 ESEC = 58496
12:47:45 SEQ = 58624 ESEC = 58368
12:47:46 SEQ = 58496 ESEC = 58752
12:47:46 SEQ = 58880 ESEC = 58624
12:47:46 SEQ = 58752 ESEC = 59008
12:47:46 SEQ = 59136 ESEC = 58880
12:47:47 SEQ = 59008 ESEC = 59264
12:47:47 SEQ = 59392 ESEC = 59136
12:47:47 SEQ = 59264 ESEC = 59520

每个 128 RTP 包都会引发错误，如您所见 |ESEC - SEQ| = 256，我还删除了 RTP header 并将数据写入 file.mpg 和视频播放正常，没有错误和伪影。

请解释我如何正确计算多播流中的 RTP 序列号

Answer 1

由于缓冲区是 signed char 数组，您的转换(左移)发生了溢出。是UB

您可以使用此代码示例测试/解决它:

#include <stdio.h>
#include <stdint.h>

int main()
{
    uint16_t seq1;
    uint16_t seq2;
    uint16_t seq3;

             char buffer1[] = {0, 0, 0xE8, 0x80};
    unsigned char buffer2[] = {0, 0, 0xE8, 0x80};

    seq1 = (buffer1[2] << 8)+buffer1[3];
    printf("seq1: %d\n", seq1);

    seq2 = (buffer2[2] << 8)+buffer2[3];
    printf("seq2: %d\n", seq2);

    seq3 = buffer1[2];
    seq3 = (seq3 << 8)+buffer2[3];
    printf("seq3: %d\n", seq3);

    return 0;
}

输出是:

seq1: 59264
seq2: 59520
seq3: 59520

如您所见，使用 signed char (seq1) 缓冲区转换是错误的。该示例显示了您发布的最后一行结果发生了什么。