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通过将数据块读入内存并使用下面显示的函数交换大端整数，我设法解析了一个大型二进制文件（~8Gb）。但是，我试图通过使用Boost Memory-Mapped 文件来获得更多性能，但我无法使用 endian_swap 函数，因为该文件是以只读模式打开的。有没有什么有效的方法可以在不写入原始文件的情况下交换字节？如果不是，性能会受到 I/O 开销的影响吗？

inline void endian_swap(unsigned short int& x)
{
  x = (x>>8) |
    (x<<8);
}
inline void endian_swap(unsigned int& x)
{
  x = (x>>24) |
    ((x<<8) & 0x00FF0000) |
    ((x>>8) & 0x0000FF00) |
    (x<<24);
}
inline void endian_swap(unsigned long long int& x)
{
  x = (((unsigned long long int)(x) << 56) | \
      (((unsigned long long int)(x) << 40) & 0xff000000000000ULL) | \
      (((unsigned long long int)(x) << 24) & 0xff0000000000ULL) | \
      (((unsigned long long int)(x) << 8)  & 0xff00000000ULL) | \
      (((unsigned long long int)(x) >> 8)  & 0xff000000ULL) | \
      (((unsigned long long int)(x) >> 24) & 0xff0000ULL) | \
      (((unsigned long long int)(x) >> 40) & 0xff00ULL) | \
      ((unsigned long long int)(x)  >> 56));
}

代码在这篇文章中找到。非常感谢您的宝贵时间

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1 回答 1

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至少底层操作系统支持您想要的行为：

   MAP_PRIVATE
              Create a private copy-on-write mapping.  Updates
              to the mapping are not visible to other processes
              mapping the same file, and are not carried through
              to the underlying file.  It is unspecified whether
              changes made to the file after the mmap() call are
              visible in the mapped region.

该priv标志似乎转换为MAP_PRIVATE：

void* data = 
    ::BOOST_IOSTREAMS_FD_MMAP( 
        const_cast<char*>(p.hint), 
        size_,
        readonly ? PROT_READ : (PROT_READ | PROT_WRITE),
        priv ? MAP_PRIVATE : MAP_SHARED,
        handle_, 
        p.offset );
if (data == MAP_FAILED)
    cleanup_and_throw("failed mapping file");
于 2011-07-03T01:30:33.163 回答