How do I convert a Python float to a hexadecimal string in python 2.5? Nonworking solution attached

Question

What I really need to do is to export a floating point number to C with no precision loss.

I did this in python:

import math
import struct
x = math.sqrt(2)
print struct.unpack('ii', struct.pack('d', x))
# prints (1719614413, 1073127582)

And in C I try this:

#include <math.h>
#include <stdio.h>

int main(void)
{
  unsigned long long x[2] = {1719614413, 1073127582};
  long long lx;
  double xf;

  lx = (x[0] << 32) | x[1];
  xf = (double)lx;
  printf("%lf\n", xf);
  return 0;
}

But in C I get:

7385687666638364672.000000 and not sqrt(2).

What am I missing?

Thanks.

Answer 1

The Python code appears to work. The problem is in the C code: you have the long long filled out right, but then you convert the integer value directly into floating point, rather than reinterpreting the bytes as a double. If you throw some pointers/addressing at it it works:

jkugelman$ cat float.c
#include <stdio.h>

int main(void)
{
    unsigned long x[2] = {1719614413, 1073127582};
    double d = *(double *) x;

    printf("%f\n", d);
    return 0;
}
jkugelman$ gcc -o float float.c 
jkugelman$ ./float 
1.414214

Notice also that the format specifier for double (and for float) is %f, not %lf. %lf is for long double.

Answer 2

If you're targeting a little-endian architecture,

>>> s = struct.pack('<d', x)
>>> ''.join('%.2x' % ord(c) for c in s)
'cd3b7f669ea0f63f'

if big-endian, use '>d' instead of <d. In either case, this gives you a hex string as you're asking for in the question title's, and of course C code can interpret it; I'm not sure what those two ints have to do with a "hex string".