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 hostshort);
uint32_t ntohl(uint32_t netlong);
uint16_t ntohs(uint16_t netshort);

functions and they worked well but none of them does double (float) conversion because these types are different on every architecture. And through the XDR i found double-float precision format representations (http://en.wikipedia.org/wiki/Double_precision) but no byte ordering there.

So, I would much appreciate if somebody helps me out on this (C code would be great!). NOTE: OS is Linux kernel (2.6.29), ARMv7 CPU architecture.

There is no network byte ordering for doubles because there is no single defined network representation for them. You can break them into a mantissa, sign and exponent if you really wanted or represent them as a numerator and a denominator but that's fiddly. — Flexo, May 16 '12 at 10:47

jeb · Accepted Answer · 2012-05-19T14:20:00.573

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);
}

I found this but receiving unexpected value on server. 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); } — , May 17 '12 at 05:13

score 1 · Answer 2 · answered Jan 30 '14 at 20:23

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.

how to convert double between host and network byte order?

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