Displaying 0 in loop

Question

Am trying to create simple while loop until my variable will be 0. So am created a

do
{
x = x + 0.001;
cout << x0 << endl;
} while (SOME_CALCULATION != 0);

Everything works great until my loop reach 0 then i getting 7.63278e-017 instead of 0.

Loop example:

result: -0.003
result: -0.002
result: -0.001
result: 7.63278e-017
result: 0.001
result: 0.002
result: 0.003

That's normal and expected behaviour of floating-point arithmetic. — gnasher729, Apr 24 '14 at 17:23
Related: [Most effective way for float and double comparison](http://stackoverflow.com/questions/17333/most-effective-way-for-float-and-double-comparison) — Shafik Yaghmour, Apr 24 '14 at 17:24
Can we assume that the use of two variables (`x` and `x0`), and the differences between the `cout` line and the reported results, in the fragment above are just typos? — Bob Jarvis - Слава Україні, Apr 24 '14 at 17:29
Not a duplicate but certainly related: http://stackoverflow.com/questions/13698927/compare-double-to-zero-using-epsilon — πάντα ῥεῖ, Apr 24 '14 at 17:40

0x90 · Answer 1 · 2014-09-30T04:20:29.847

3

Do something like that:

    double epsilon = 0.00001; 
    do
    {
        x = x + 0.001;
        cout << x0 << endl;
    } while (abs(SOME_CALCULATION) > epsilon);

edited Sep 30 '14 at 04:20

answered Apr 24 '14 at 17:25

0x90

39,472
36
165
245

score 2 · Answer 2 · answered Apr 24 '14 at 17:33

You might want to consider rewriting this as

x = -3;

do
  {
  x = x + 1;
  cout << x0 / 1000 << endl;
  } while (SOME_CALCULATION != 0);

SOME_CALCULATION will need to be adjusted for the fact that x is now an integer rather than a fractional floating point value - perhaps replicating the x0 / 1000 within the calculation would work.

Share and enjoy.

MooseBoys · Answer 3 · 2014-04-24T20:35:53.450

Some real numbers cannot be represented exactly in floating point (for example, 1.0/3.0). As a result, rounding errors are introduced, and can accumulate in surprising ways. Check out this link for more information.

Depending on your algorithm, a simple workaround could be to check for <=0 rather than ==0. In general, if you find yourself checking equivalency of a floating point value, you should think twice and ensure that all possible values are exactly representable using the type specified.

score 0 · Answer 4 · answered Apr 24 '14 at 17:36

Workaround: Used long or int variables and convert to double before output:

#include <iostream>

using namespace std;

main()
{
    int x=-3;
    do
    {
        x = x + 1;
        cout << ((double)x)/1000 << endl;
    } while (x < 4);
}

Output:

-0.002
-0.001
0
0.001
0.002
0.003
0.004

Displaying 0 in loop

4 Answers4