这里有一个基本的函数来交换大小端序。它是基本的，但不需要补充库。

void endianness_swap(uint32_t& val) {
    uint8_t a, b, c;
    a = (val & 0xFF000000) >> 24;
    b = (val & 0x00FF0000) >> 16;
    c = (val & 0x0000FF00) >> 8;
    val=(val & 0x000000FF) << 24;
    val = val + (c << 16) + (b << 8) + (a);
}

查找位移位，因为这基本上是所有你需要做的交换从小->大端dian。然后根据位的大小，改变位移位的方式。

void writeLittleEndianToBigEndian(void* ptrLittleEndian, void* ptrBigEndian , size_t bufLen )
{
    char *pchLittleEndian = (char*)ptrLittleEndian;

    char *pchBigEndian = (char*)ptrBigEndian;

    for ( size_t i = 0 ; i < bufLen ; i++ )    
        pchBigEndian[bufLen-1-i] = pchLittleEndian[i];
}

std::uint32_t row = 0x12345678;

char buf[4]; 

writeLittleEndianToBigEndian( &row, &buf, sizeof(row) );

和在C中一样:

short big = 0xdead;
short little = (((big & 0xff)<<8) | ((big & 0xff00)>>8));

您还可以声明一个无符号字符的向量，将输入值memcpy放入其中，将字节反向转换为另一个向量，然后将字节memcpy取出，但这将花费比旋转位长几个数量级的时间，特别是对于64位值。

我们已经用模板做到了这一点。你可以这样做:

// Specialization for 2-byte types.
template<>
inline void endian_byte_swapper< 2 >(char* dest, char const* src)
{
    // Use bit manipulations instead of accessing individual bytes from memory, much faster.
    ushort* p_dest = reinterpret_cast< ushort* >(dest);
    ushort const* const p_src = reinterpret_cast< ushort const* >(src);
    *p_dest = (*p_src >> 8) | (*p_src << 8);
}

// Specialization for 4-byte types.
template<>
inline void endian_byte_swapper< 4 >(char* dest, char const* src)
{
    // Use bit manipulations instead of accessing individual bytes from memory, much faster.
    uint* p_dest = reinterpret_cast< uint* >(dest);
    uint const* const p_src = reinterpret_cast< uint const* >(src);
    *p_dest = (*p_src >> 24) | ((*p_src & 0x00ff0000) >> 8) | ((*p_src & 0x0000ff00) << 8) | (*p_src << 24);
}

我喜欢这个，只是为了风格:-)

long swap(long i) {
    char *c = (char *) &i;
    return * (long *) (char[]) {c[3], c[2], c[1], c[0] };
}