Modulus with double variables

[Peter Pippinger]

Hello NG,

i need to calculate the modulus with double precisson typed variables.

this works well with integers:
w1 = w1 % 360;

but what do i have to do if i want to use w1 as double?

Thanks a lot!
Peter

 

[Jim Langston]

Hello NG,

i need to calculate the modulus with double precisson typed variables.

this works well with integers:
w1 = w1 % 360;

but what do i have to do if i want to use w1 as double?

Well, you have the problem that floating point division does not have a
remainder, otherwise it wouldn't be floating point. But, there is still a
solution.

Take for an example
14.5 / 3.5;
The answer is:
4.1428571428571428571428571428571
So what would the remainder be defined as? This would be how many times 3.5
goes into 14.5 wholy. We see it goes 4 times plus a little bit. This is
easy to figure out. Take the answer as an integer. Multiply it by the
original divider. Then subtract this by the original divider.

float a = 14.5;
float b = 3.5;
int result = static_cast<int>( a / b );
float mod = a - static_cast<float>( result ) * b;

 

[Peter Pippinger]

wow! That´s great. I have packed this in a function:

//
---------------------------------------------------------------------
// -- Modulus mit double Variablen
//
---------------------------------------------------------------------
double modulus(double a, double b)
{
int result = static_cast<int>( a / b );
return a - static_cast<double>( result ) * b;
}

Thanks a lot! you are great!

 

[Eric Jensen]

Hello NG,

i need to calculate the modulus with double precisson typed variables.

this works well with integers:
w1 = w1 % 360;

but what do i have to do if i want to use w1 as double?

Thanks a lot!
Peter

double fmod ( double x, double y ); // #include <cmath>

double r,x=5.3,y=2;
r = fmod(x,y); // r is not equal to 1.3

//eric

 

[Alf P. Steinbach]

* Peter Pippinger:

i need to calculate the modulus with double precisson typed variables.

this works well with integers:
w1 = w1 % 360;

but what do i have to do if i want to use w1 as double?

Check out round and floor.

 

[Steve Pope]

float a = 14.5;
float b = 3.5;
int result = static_cast<int>( a / b );
float mod = a - static_cast<float>( result ) * b;

I would worry that the static_cast would round to the
nearest integer rather than truncating towards negative
infinity.

Steve