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  1. #include <time.h>
  2.  
  3. #include <iostream>
  4. using namespace std;
  5.  
  6. int main() {
  7. 	// your code goes here
  8. 	double duration = 24000.0/1001.0;
  9. 	double first_pts = (double)time(NULL);
  10. 	unsigned long long nb_frame = 1000000;
  11.  
  12. 	double time_after_1 = first_pts + nb_frame * duration;
  13. 	double time_after_2 = first_pts;
  14. 	for(unsigned long long i = 0 ; i < nb_frame ; ++i)
  15. 	{
  16. 		time_after_2 += duration;
  17. 	}
  18.  
  19. 	double time_after_3 = first_pts;
  20. 	double c = 0.0;             // A running compensation for lost low-order bits.
  21.     for(unsigned long long i = 0 ; i < nb_frame ; ++i)
  22.     {
  23.     	double y = duration - c;    // So far, so good: c is zero.
  24.         double t = time_after_3 + y;          // Alas, sum is big, y small, so low-order digits of y are lost.
  25.         c = (t - time_after_3) - y; // (t - sum) recovers the high-order part of y; subtracting y recovers -(low part of y)
  26.         time_after_3 = t;           // Algebraically, c should always be zero. Beware overly-aggressive optimizing compilers!
  27.         // Next time around, the lost low part will be added to y in a fresh attempt.
  28.     }
  29. 	cout << time_after_1 - time_after_2 << " or " << time_after_1 - time_after_3 << endl;
  30.  
  31. 	return 0;
  32. }
Success #stdin #stdout 0.01s 3340KB
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3.14042e-05 or 1.74623e-09
https://ideone.com/RQmW15
an example of floating point error accumulation
language:
C++14 (gcc 8.3)
created:
10 years ago
visibility:
public

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