how to convert double between host and network byte order?

Question

Could somebody tell me how to convert double precision into network byte ordering. I tried

uint32_t htonl(uint32_t hostlong);
uint16_t htons(uint16_t hostsh         


        

Answer 1

This might be hacky (the char* hack), but it works for me:

double Buffer::get8AsDouble(){

    double little_endian = *(double*)this->cursor;

    double big_endian;

    int x = 0;
    char *little_pointer = (char*)&little_endian;
    char *big_pointer = (char*)&big_endian;

    while( x < 8 ){
        big_pointer[x] = little_pointer[7 - x];
        ++x;
    }

    return big_endian;

}

For brevity, I've not include the range guards. Though, you should include range guards when working at this level.

Answer 2

You could look at IEEE 754 at the interchanging formats of floating points.

But the key should be to define a network order, ex. 1. byte exponent and sign, bytes 2 to n as mantissa in msb order.

Then you can declare your functions

uint64_t htond(double hostdouble);
double ntohd(uint64_t netdouble);

The implementation only depends of your compiler/plattform.
The best should be to use some natural definition, so you could use at the ARM-platform simple transformations.

EDIT:

From the comment

static void htond (double &x)
{
  int *Double_Overlay; 
  int Holding_Buffer; 
  Double_Overlay = (int *) &x; 
  Holding_Buffer = Double_Overlay [0]; 
  Double_Overlay [0] = htonl (Double_Overlay [1]); 
  Double_Overlay [1] = htonl (Holding_Buffer); 
}

This could work, but obviously only if both platforms use the same coding schema for double and if int has the same size of long.
Btw. The way of returning the value is a bit odd.

But you could write a more stable version, like this (pseudo code)

void htond (const double hostDouble, uint8_t result[8])
{
  result[0] = signOf(hostDouble);
  result[1] = exponentOf(hostDouble);
  result[2..7] = mantissaOf(hostDouble);
}