[置顶] HEVC学习（十七） —— NAL unit 的解码过程之一

下图为官方标准中NAL层的句法元素，且以伪代码的形式给出了解码过程：

在HM中由TAppDecTop::decode()调用byteStreamNALUnit(bytestream, nalUnit, stats)实现如上伪代码：

/**
 * Parse an AVC AnnexB Bytestream bs to extract a single nalUnit
 * while accumulating bytestream statistics into stats.
 *
 * Returns false if EOF was reached (NB, nalunit data may be valid),
 *         otherwise true.
 */
Bool
byteStreamNALUnit(
  InputByteStream& bs,
  vector<uint8_t>& nalUnit,
  AnnexBStats& stats)
{
  Bool eof = false;
  try
  {
    _byteStreamNALUnit(bs, nalUnit, stats); //!< 实际完成NAL解析工作的函数
  }
  catch (...) //!< 捕获所有异常
  {
    eof = true;
  }
  stats.m_numBytesInNALUnit = UInt(nalUnit.size());
  return eof;
}

在分析NAL解析过程之前，先介绍几个会被调用到的子函数，以便更好地理解解析过程。

1. Bool eofBeforeNBytes(UInt n)

如果在读码流的接下来的n字节的过程中遇到了文件结束符，则该函数返回true，否则返回false。

  /**
   * returns true if an EOF will be encountered within the next
   * n bytes.
   */
  Bool eofBeforeNBytes(UInt n)
  {
    assert(n <= 4);
    if (m_NumFutureBytes >= n) //!< m_NumFutureBytes大于等于n只会在该函数被调用2次及2次以上的情况下发生，满足该条件时无须继续读多余的字节，故返回false
      return false;

    n -= m_NumFutureBytes; //!< n先减去m_NumFutureBytes的目的是防止被函数peekBytes调用时再读入接下来的n字节数据
    try
    {
      for (UInt i = 0; i < n; i++)
      {
        m_FutureBytes = (m_FutureBytes << 8) | m_Input.get(); //!< 每次读入一个字节，循环结束后，m_FutureBytes存放的是读入的n个字节的数据
        m_NumFutureBytes++;
      }
    }
    catch (...) //!< 出现异常即读到文件结尾，返回true
    {
      return true;
    }
    return false;
  }

2. uint32_t peekBytes(UInt n)

该函数在不移动文件指针的前提下返回文件中接下来的n字节。实现的即是伪代码中的next_bits(n)的功能。

  /**
   * return the next n bytes in the stream without advancing
   * the stream pointer.
   *
   * Returns: an unsigned integer representing an n byte bigendian
   * word.
   *
   * If an attempt is made to read past EOF, an n-byte word is
   * returned, but the portion that required input bytes beyond EOF
   * is undefined.
   *
   */
  uint32_t peekBytes(UInt n)
  {
    eofBeforeNBytes(n);
    return m_FutureBytes >> 8*(m_NumFutureBytes - n); //!< 若m_NumFutureBytes=4, n=3，则返回m_FutureBytes左移8位后（即有效数据位为3字节）的数据
  }

3. uint8_t readByte()

该函数读文件的一个字节并返回。

  /**
   * consume and return one byte from the input.
   *
   * If bytestream is already at EOF prior to a call to readByte(),
   * an exception std::ios_base::failure is thrown.
   */
  uint8_t readByte()
  {
    if (!m_NumFutureBytes) //!< m_FutureBytes为NULL，则从文件中读入一个字节并返回
    {
      uint8_t byte = m_Input.get();
      return byte;
    }//! m_FutureBytes非NULL，则从它当中取出一个字节出来
    m_NumFutureBytes--; //!< 计数值减1
    uint8_t wanted_byte = m_FutureBytes >> 8*m_NumFutureBytes; //!< m_FutureBytes为4字节，取出有效数据中的最高字节
    m_FutureBytes &= ~(0xff << 8*m_NumFutureBytes); //!< 对应位置的数据清零
    return wanted_byte;
  }

4. uint32_t readBytes(UInt n)

该函数读文件的n个字节并返回。

  /**
   * consume and return n bytes from the input.  n bytes from
   * bytestream are interpreted as bigendian when assembling
   * the return value.
   */
  uint32_t readBytes(UInt n)
  {
    uint32_t val = 0;
    for (UInt i = 0; i < n; i++)
      val = (val << 8) | readByte(); //!< 每次调用readByte()读入一个字节，通过对val左移8位且与输入值进行或运算实现将n个字节存储到val这个变量中
    return val;
  }