C++

Question

我觉得我眼界有点狭隘，所以我需要你的帮助。

我正在尝试解析 MPEG Transport-stream file 但我卡在 Header 上，在 wiki 上您会看到提供了一些 32 位 BE MASK 以便从中提取数据4 个字节 header。我的代码考虑了字节序（我认为），如果它检测到你在小字节序上运行则反转字节。然后我将 char* 转换为 int 并应用掩码，所有值看起来都很好，但 PID 搞砸了，我不明白为什么...

header definition

namespace ts {

#define SYNC_BYTE_MASK 0xff000000
#define TEI_MASK 0x800000
#define PAYLOAD_START_MASK 0x400000
#define PRIORITY_MASK 0x200000
#define PID_MASK 0x1fff00
#define SCRAMBLING_CTL_MASK 0xc0
#define ADAPTATION_FIELD_MASK 0x20
#define HAS_PAYLOAD_MASK 0x10
#define COUNTER_MASK 0xf

#define HEADER_BYTES 4
#define HEADER_BITS 8 * HEADER_BYTES

class Header {

public:
    std::bitset<HEADER_BITS> *full;

    unsigned char _syncByte;
    bool _tei;
    bool _payloadStart;
    bool _priority;
    int16_t _pid;
    std::bitset<2> *_scramblingCtl;
    bool _adaptationField;
    bool _hasPayload;
    int _counter;

    Header(const char *, size_t);
    ~Header();

    const std::string toString();
    bool isValid();

};

}

Header values assignment

ts::Header::Header(const char *header, size_t n) {
    uint32_t bytes = reverseLE(header, n);

    // just for display
    char t[4];
    memcpy(t, header, 4);
    std::cout << "Original: " << std::bitset<32>(*((uint32_t *)t)) << std::endl;
    this->full = new std::bitset<HEADER_BITS>(bytes);

    uint32_t tmp = bytes & SYNC_BYTE_MASK;
    this->_syncByte = ((char *)&tmp)[n - 1];

    this->_tei = bytes & TEI_MASK;
    this->_payloadStart = bytes & PAYLOAD_START_MASK;
    this->_priority = bytes & PRIORITY_MASK;
    this->_pid = bytes & PID_MASK; // THIS ONE IS MESSED UP !!
    this->_scramblingCtl = new std::bitset<2>(bytes & SCRAMBLING_CTL_MASK);
    this->_adaptationField = bytes & ADAPTATION_FIELD_MASK;
    this->_hasPayload = bytes & HAS_PAYLOAD_MASK;
    this->_counter = bytes & COUNTER_MASK;
}

Functions to reverse

   #include "utils.h"

int is_big_endian(void)
{
    union {
        uint32_t i;
        char c[4];
    } e = { 0x01000000 };

    return e.c[0];
}

void swap(char *s, int a, int b) {
    char tmp;

    tmp = s[a];
    s[a] = s[b];
    s[b] = tmp;
}

// Converts string to int taking endianess into account
uint32_t reverseLE(const char *bits, size_t n) {
    uint32_t ret = 0;
    char *cp = (char *)malloc(n * sizeof(char));

    memcpy(cp, bits, n);
    if ( ! is_big_endian() ) {
        for (int i = 0; i < n / 2; i++)
            swap(cp, i, n - 1 - i);
    }

    ret = *((uint32_t *)cp);
    free(cp);
    return ret;
}

这是一个 header 的示例，其 PID 应为 33

Original: 00010010001000010000000001000111
Binary: 01000111000000000010000100010010
Sync byte: G
TEI: 0
Payload start: 0
Priority: 0
PID: 8448 0010000100000000
Scrambling Ctl: 00
Adaptation field: 0
Has Payload: 1
Counter: 2

不知何故又反转了，我不明白为什么...

Answer 1

好的，问题是 PID 的 13 位位于 str[1] 和 str[2]，这意味着在转换 *((int *)str) 并应用掩码后，仍然有 8 个尾随 0最后一个字节的位 str[3].

解决方案：

this->_pid = bytes & PID_MASK; 
this->_pid >>= 8;

感谢@Wimmel。

C++ - MPEG TS - 解析 header - PID 混乱 - Big endian 32 位掩码

C++ - MPEG TS - Parsing header - PID is messed up - Big endian 32 bits mask

parsing

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mpeg2-ts