如何在C ++中将little-endian 64转换为主机字节顺序

Question

I need to convert little-endian 64 to host byte order. In winapi i can't find such functions, so i need to write my own, can anyone help me? Thanks!

Answer 1

Use htonll . It converts an unsigned __int64 to network byte order.

Answer 2

I think you need to get the host's endiannes first and you can decide after that if you need to convert anything:

#define BIG_ENDIAN 1
#define LITTLE_ENDIAN 0

int getEndiannes()
{
   int n = 1;
   char *p = &n;
   if(*p) 
       return LITTLE_ENDIAN; 
   else
       return BIG_ENDIAN ;  

}

Answer 3

In Linux you can use uint64_t htobe64(uint64_t host_64bits);

Check the man page for more details.

Answer 4

If you're reading external data, the usual solution is to build up the individual values as specified:

unsigned long long      //  The only type guaranteed long enough for 64 bits
readData( std::istream& source )
{
    unsigned long long results = source.get();
    results |= source.get() <<  8;
    results |= source.get() << 16;
    results |= source.get() << 24;
    results |= source.get() << 32;
    results |= source.get() << 40;
    results |= source.get() << 48;
    results |= source.get() << 56;
    return results;
}

Of course, you really need some sort of error checking, in case the file ends in the middle of the 8 bytes. (But it is sufficient to check once, after all of the bytes have been read.)

If the data is already in a buffer, then just substitute static_cast<unsigned char>(*p++) for source.get() (where p points to the position in the buffer). In this case, you also have to ensure that there are 8 bytes between the initial p and the end of the buffer before doing the conversion.