Сounter of digits after decimal point (double)

Question

Pls, help with this problem. I need to count the number of digits after the decimal point. But when f should be 0 it becomes 1 and i becomes 2 (but should be 3). After the code, the output from the console

#include <math.h>
#include <stdio.h>       

int main()
{
  double i;
  double f;
  int count = 0;
  double eps = 10.123;

  while(f!=0){
  f = modf(eps, &i);
  printf("%f \n%f\n", i, f);
  count = count + 1;
  i=0;
  eps=f*10;
  printf("%f\n\n", eps);
  }
  printf("Count = %d", count);

    return 0;
}

10.000000 0.123000 1.230000

1.000000 0.230000 2.300000

2.000000 0.300000 3.000000

2.000000 1.000000 10.000000

9.000000 1.000000 10.000000

9.000000 1.000000 9.999999

9.000000 0.999999 9.999993

9.000000 0.999993 9.999933

9.000000 0.999933 9.999332

9.000000 0.999332 9.993321

9.000000 0.993321 9.933209

9.000000 0.933209 9.332090

9.000000 0.332090 3.320898

3.000000 0.320898 3.208983

3.000000 0.208983 2.089828

2.000000 0.089828 0.898284

0.000000 0.898284 8.982839

8.000000 0.982839 9.828386

9.000000 0.828386 8.283855

8.000000 0.283855 2.838551

2.000000 0.838551 8.385506

8.000000 0.385506 3.855058

3.000000 0.855058 8.550583

8.000000 0.550583 5.505825

5.000000 0.505825 5.058253

5.000000 0.058253 0.582528

0.000000 0.582528 5.825281

5.000000 0.825281 8.252811

8.000000 0.252811 2.528114

2.000000 0.528114 5.281143

5.000000 0.281143 2.811432

2.000000 0.811432 8.114319

8.000000 0.114319 1.143188

1.000000 0.143188 1.431885

1.000000 0.431885 4.318848

4.000000 0.318848 3.188477

3.000000 0.188477 1.884766

1.000000 0.884766 8.847656

8.000000 0.847656 8.476562

8.000000 0.476562 4.765625

4.000000 0.765625 7.656250

7.000000 0.656250 6.562500

6.000000 0.562500 5.625000

5.000000 0.625000 6.250000

6.000000 0.250000 2.500000

2.000000 0.500000 5.000000

5.000000 0.000000 0.000000

Count = 50

10.123 is actually : `10.1229999999999993320898283855058252811431884765625` https://godbolt.org/z/7Tha1cor9 — 0___________, Nov 02 '21 at 14:19

score 0 · Accepted Answer · answered Nov 02 '21 at 14:17

0

Just because you assigned 10.123 to the variable does not mean that is exactly what got stored there.

Floating point representations are only approximate (although they do have very high precision)

10.123 cannot be precisely stored, and the binary approximation is 10.1229999999..., as shown by this code.

#include <stdio.h>

int main(void) {
    double x = 10.123;
    
    printf("%3.3f is stored as %3.20f\n", x, x);
    return 0;
}

Output:

10.123 is stored as 10.12299999999999933209

In otherwords, your code is working properly.

answered Nov 02 '21 at 14:17

abelenky

63,815
23
109
159

rather `10.1229999999999993320898283855058252811431884765625` – 0___________ Nov 02 '21 at 14:21
Is there any way to count the number of digits after the decimal point? – Anastay Kaiser Nov 02 '21 at 14:22
Yes. Do it as a string before creating the inexact floating point variable. – stark Nov 02 '21 at 14:52

Сounter of digits after decimal point (double)

1 Answers1

Output:

In otherwords, your code is working properly.