Re: conversion problem int <-> double ?

Jack Klein <>
Mon, 31 Mar 2008 21:44:44 -0500
On Mon, 31 Mar 2008 19:04:01 -0700 (PDT), Markus Dehmann
<> wrote in comp.lang.c++:

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;

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 precision.

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

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

Your basic idea is wrong.


#include <iostream>

void decode(double n, int& i1, int& i2){
  i1 = int(n);
  double rest = n - int(n);
  i2 = int(rest * 100.0); // i2 is 1, should be

int main(int argc, char** argv){
  double n = 324.02;
  int p;
  int i;
  decode(n, p, i);
  std::cerr << "n=" << n <<", p=" << p << ", i=" << i << std::endl;
  return EXIT_SUCCESS;

Look at this short program:

#include <iostream>
#include <iomanip>

int main()
   double d = 304.0;
   d += (2 / 100.0);

   std::cout << "The value is " << std::setprecision(20) << d <<
   return 0;

Here is the output of that program on my computer:

The value is 304.01999999999998

If you can't think of any better idea than trying to stick two integer
values into a double, and there is almost certainly a better way, here
are a few possible approaches:

1. Since one number is always between 0 and 99, you could multiply
the other number by 100 and add the second one. This will work if the
first value is not too large to fit into a double when multiplied by
100. You can calculate this by using the value of the macro DBL_DIG
in the <cfloat> or <float.h> header.

In my implementation, this value is 15, which means that a double can
hold a whole number value up to 999,999,999,999,999 with no loss of
precision. So if the first number is guaranteed not to be greater
than 1/100 of this value, approach 1 will work.

2. If you must stick integer values into floating point fractions, do
not simply multiply them back up and assign them to an int. Assignment
to an integer type causes truncation, any fractional portion is just
chopped off. So .01999999999998 * 100 equals 1.999999999998 which
gets truncated to 1.

Instead, if you know the fraction is positive, pass it to the
std::ceil() function before converting to int.

Jack Klein
Home: http://JK-Technology.Com
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