Thread: Portable Doubles

Originally Posted by siavoshkc

I don't understand why. What is the 400921FB54442D18? Hexadecimal of 3.14...? Why its character string uses 8 bytes?

It's a case of people making remarks out of context. The two points are:

1)3.14... as a double takes up just as much space if you print in in hex.
2)as a character string, printing hex is shorter than desimal in this particular case.

Originally Posted by EvilGuru

does anyone known of a smarter method to swap the bytes than this:

Code:

void my_htond (double *out, double in)
{
    unsigned char *bin = (unsigned char *) ∈
    unsigned char *bot = (unsigned char *) out;
    int j;
    for (j = 0; j < sizeof(double); j++) bout[j] = bin[(sizeof(double) - 1) - j];
}

This is pretty much the same thing, except done in place -- just for something else to look at.

Code:

#include <stdio.h>

void reverse_bytes(void *object, size_t size)
{
   unsigned char *beg, *end;
   for ( beg = object, end = beg + size - 1; beg < end; ++beg, --end )
   {
      unsigned char temp = *beg;
      *beg = *end;
      *end = temp;
   }
}

void show_bytes(const void *object, size_t size)
{
   const unsigned char *byte;
   for ( byte = object; size--; ++byte )
   {
      printf("%p : %02X\n", byte, (unsigned)*byte);
   }
}

int main(void)
{
   double value = 123.456;
   puts("a:");
   show_bytes(&value, sizeof value);
   puts("b:");
   reverse_bytes(&value, sizeof value);
   show_bytes(&value, sizeof value);
   return 0;
}

/*
http://babbage.cs.qc.edu/courses/cs341/IEEE-754.html:
405EDD2F1A9FBE77
*/

/* my output
a:
0012FF84 : 77
0012FF85 : BE
0012FF86 : 9F
0012FF87 : 1A
0012FF88 : 2F
0012FF89 : DD
0012FF8A : 5E
0012FF8B : 40
b:
0012FF84 : 40
0012FF85 : 5E
0012FF86 : DD
0012FF87 : 2F
0012FF88 : 1A
0012FF89 : 9F
0012FF8A : BE
0012FF8B : 77
*/