Re: how to encode a float in base64?

From:
James Kanze <james.kanze@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Mon, 17 Mar 2008 03:26:51 -0700 (PDT)
Message-ID:
<ac959561-debd-42ff-938a-4d13908f9762@e39g2000hsf.googlegroups.com>
On Mar 17, 9:01 am, Paul Brettschneider <paul.brettschnei...@yahoo.fr>
wrote:

Jack Klein wrote:


    [...]

Then you can memcpy() the float into the beginning of the
buffer. Then you can encode the 6 byte buffer.


Isn't using unions the idiomatic thing to do in this case?


It's formally undefined behavior (although I suspect that most
compilers support it). The standard sanctionned way is with
reinterpret_cast, but this is not without problems, and is
probably less portable than the union in practice. memcpy is
guaranteed to work, everywhere (and also avoids any alignment
issues which might otherwise crop up).

Like this:

#include <iostream>
#include <cmath>
#include <algorithm>
#include <iterator>

int main()
{
        const size_t size = sizeof(double);
        union {
                double f;
                unsigned char s[size];
        } u;
        u.f = 4.0 * std::atan(1.0); // Pi
        std::cout << u.f << '\n';
        std::copy(&u.s[0], &u.s[size],
                  std::ostream_iterator<unsigned int>(std::cout, "-"));
        std::cout << std::endl;

}

On IA32:
3.14159
24-45-68-84-251-33-9-64-

On PA-RISC:
3.14159
64-9-33-251-84-68-45-24-

;)


On a Sun Sparc, if I try this somewhere in the middle of a
larger buffer (say at the second byte), I get a core dump:-).
(Correctly aligned, or using memcpy, the results are the same as
those of the PA-RISC.) Try it on just about any mainframe, and
you'll get still other values.

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James Kanze (GABI Software) email:james.kanze@gmail.com
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