How to byteswap a double?

Try 3

Okay, found out there's a better way. The other way you have to worry about the order you pack/unpack stuff. This way you don't:

// int and float
static void swap4(void *v)
{
    char    in[4], out[4];
    memcpy(in, v, 4);
    out[0] = in[3];
    out[1] = in[2];
    out[2] = in[1];
    out[3] = in[0];
    memcpy(v, out, 4);
}

// double
static void swap8(void *v)
{
    char    in[8], out[8];
    memcpy(in, v, 8);
    out[0] = in[7];
    out[1] = in[6];
    out[2] = in[5];
    out[3] = in[4];
    out[4] = in[3];
    out[5] = in[2];
    out[6] = in[1];
    out[7] = in[0];
    memcpy(v, out, 8);
}

typedef struct
{
    int theint;
    float   thefloat;
    double  thedouble;
} mystruct;


static void swap_mystruct(void *buf)
{
    mystruct    *ps = (mystruct *) buf;
    swap4(&ps->theint);
    swap4(&ps->thefloat);
    swap8(&ps->thedouble);
}    

Send:

    char    buf[sizeof (mystruct)];
    memcpy(buf, &s, sizeof (mystruct));
    swap_mystruct(buf);

Recv:

    mystruct    s;
    swap_mystruct(buf);
    memcpy(&s, buf, sizeof (mystruct));

Although a double in main memory is 64 bits, on x86 CPUs double-precision registers are 80 bits wide. So if one of your values is stored in a register throughout, but the other makes a round-trip through main memory and is truncated to 64 bits, this could explain the small differences you're seeing.

Maybe you can force variables to live in main memory by taking their address (and printing it, to prevent the compiler from optimizing it out), but I'm not certain that this is guaranteed to work.

    b = byteswap(a);

That's a problem. After swapping the bytes, the value is no longer a proper double. Storing it back to a double is going to cause subtle problems when the FPU normalizes the value. You have to store it back into an __int64 (long long). Modify the return type of the method.