__device__ double remainder ( double  x,
double  y 
)

Compute double-precision floating-point remainder r of dividing x by y for nonzero y. Thus $ r = x - n y$. The value n is the integer value nearest $ \frac{x}{y} $. In the case when $ | n -\frac{x}{y} | = \frac{1}{2} $, the even n value is chosen.

Returns:
  • remainder(x, 0) returns NaN.
  • remainder($\pm \infty$, y) returns NaN.
  • remainder(x, $\pm \infty$) returns x for finite x.
Note:
For accuracy information for this function see the CUDA C Programming Guide, Appendix C, Table C-2.


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