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fmod, fmodf

Visual Studio 2010

Calculates the floating-point remainder.

Copy 
double fmod( 
   double x,
   double y 
);
float fmod(
   float x,
   float y 
);  // C++ only
long double fmod(
   long double x,
   long double y
);  // C++ only
float fmodf( 
   float x,
   float y 
);
x, y

Floating-point values.

fmod returns the floating-point remainder of x / y. If the value of y is 0.0, fmod returns a quiet NaN. For information about representation of a quiet NaN by the printf family, see printf.

The fmod function calculates the floating-point remainder f of x / y such that x = i * y + f, where i is an integer, f has the same sign as x, and the absolute value of f is less than the absolute value of y.

C++ allows overloading, so you can call overloads of fmod. In a C program, fmod always takes two doubles and returns a double.

Function

Required header

fmod , fmodf

<math.h>

For additional compatibility information, see Compatibility in the Introduction.

Copy 
// crt_fmod.c
// This program displays a floating-point remainder.

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

int main( void )
{
   double w = -10.0, x = 3.0, z;

   z = fmod( w, x );
   printf( "The remainder of %.2f / %.2f is %f\n", w, x, z );
}

The remainder of -10.00 / 3.00 is -1.000000
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.NET equivalent
In porting a C++ project to C#, I discovered that IEEERemainder, the supposed the ".NET Equivalent" mentioned above, is not really equivalent at all. Specifically, as best as I can tell, fmod() would need to be implemented in .NET as follows:


private double fmod(double x, double y)
{
double remainder = Math.IEEERemainder(x, y);
if (!Double.IsNaN(remainder) && remainder != 0.0)
{
if (x >= 0.0)
{
if (remainder < 0.0)
remainder += Math.Abs(y);
}
else // x < 0.0
{
if (remainder > 0.0)
remainder -= Math.Abs(y);
}
}
return remainder;
}


Hope this helps someone else out there.
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