Re: conversion problem int <-> double ?

James Kanze <>
Tue, 1 Apr 2008 01:10:24 -0700 (PDT)
On Apr 1, 4:44 am, Jack Klein <> wrote:

On Mon, 31 Mar 2008 19:04:01 -0700 (PDT), Markus Dehmann
<> wrote in comp.lang.c++:

    [Just a few odd comments, since the basic problem has
    already been addressed...]

I have two integers i1 and i2, the second of which is
guaranteed to be between 0 and 99, and I encode them into
one double:

double encoded = (double)i1 + (double)i2 / (double)100;

The obvious question is: why? If you really do receive a value
in this format (i.e. integer part and hundredths as two separate
values), and want to treat it as a single value, fine, but then
I don't understand why you want to go the other direction later.
And I can't think of any other reason why one would want to do

This is not a very good idea, as you have found. The floating
point data types in C, and just about every other computer
language, use a fixed number of bits, and that limits their

In particular, the floating point representation in almost all
computer systems, and certainly in all common ones programmed
in C++, use binary fractions.

At least one architecture that is relatively common (IBM
mainframes) used base 16, and there's at least one base 8 out
there still being sold, but that doesn't change anything---all
of your comments which follow apply to any base which is a power
of 2.

So does anyone know of a machine for which there existed a C++
compiler (or even a C compiler) which doesn't use a base which
is a power of 2. I know that machines using base 10 existed in
the past, but the ones I know of were out of production long
before even C came along. Or maybe there is a compiler for IBM
mainframes which uses their decimal arithmetic, rather than
their floating point, for float and double (but I'd be very

That means a value like .125, or .25, or .5 is exactly
representable in the fractional part of floating point values,
but fractions that are not 1/(a power of 2) are not. They get
rounded to the nearest binary fraction.

Just a nit, but that should be fractions that are not n/(a power
of 2), where n is an integer. Something like .75 is no problem
either. (Of course, if the power of 2 is greater than something
like 51, you might get problems with some of those as well.)

So, for example, 324 and 2 become 324.02. Now I want to
decode them using the function given below but it decodes
the example as 324 and 1, instead of 324 and 2.

Actually, it does not become 324.02, it becomes some value
slightly greater or smaller than 324.02, because .02 cannot be
exactly represented in a binary fraction.

Can anyone tell me what's wrong and how to do this right?
(my code see below)

Your basic idea is wrong.

Hard to say without really knowing what his basic idea is:-).
Why does he want to do this? Anyway, two "obvious" solutions
come to mind:

 -- pass through a textual representation:

        std::ostringstream s1 ;
        s1.precision( 2 ) ;
        s1.setf( std::ios::fixed, std::ios::floatfield ) ;
        s1 << encoded ;
        std::istringstream s2( s1.str() ) ;
        char dummyForDecimal ;
        s1 >> i1 >> dummyForDecimal >> i2 ;

 -- use the correct functions from C:

        double i1d ;
        i2 = nearbyint( 100.0 * modf( encoded, &i1d ) ) ;
        i1 = i1d ;

Modf is in C90, and thus in C++ (in <cmath>). Nearbyint is an
addition of C99, and thus will be in the next version of C++,
and is possibly already available in your current C++ compiler.
If not, replace the line with
        i2 = floor( 100.0 * modf( encoded, &i1d ) + 0.5 ) ;
Although less robust, it should work for positive values
constructed as above.

James Kanze (GABI Software)
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