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.s         


        

Answer 1

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".

Answer 2

repr() is your friend.

C:\junk\es2>type es2.c
#include <stdio.h>
#include <math.h>
#include <assert.h>

int main(int argc, char** argv) {
    double expected, actual;
    int nconv;
    expected = sqrt(2.0);
    printf("expected: %20.17g\n", expected);
    actual = -666.666;
    nconv = scanf("%lf", &actual);
    assert(nconv == 1);
    printf("actual:   %20.17g\n", actual);
    assert(actual == expected);
    return 0;
    }


C:\junk\es2>gcc es2.c

C:\junk\es2>\python26\python -c "import math; print repr(math.sqrt(2.0))" | a
expected:   1.4142135623730951
actual:     1.4142135623730951

C:\junk\es2>

Answer 3

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.