C 从多播流中计算RTP序列号
我试图从多播流中计算RTP序列号并确定丢失的包 请参阅以下C代码:C 从多播流中计算RTP序列号,c,sockets,multicast,rtp,C,Sockets,Multicast,Rtp,我试图从多播流中计算RTP序列号并确定丢失的包 请参阅以下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,
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;
}
}
}
每个128
RTP包都会出现错误,正如您所看到的|ESEC-SEQ |=256
,我还删除了RTP头并将数据写入文件.mpg,视频播放良好,没有错误和伪影。
请解释如何正确计算多播流中的RTP序列号由于缓冲区是
有符号字符
数组,您的转换(左移)溢出。是的
您可以使用以下代码示例测试/解决此问题:
#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;
}
正如您可以看到的,使用有符号字符(seq1
)缓冲区时,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。它不是代码。此(缓冲区[2]
#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;
}
seq1: 59264
seq2: 59520
seq3: 59520