Bhaxy
Bhaxy

Reputation: 5574

Can't use modulus on doubles?

I have a program in C++ (compiled using g++). I'm trying to apply two doubles as operands to the modulus function, but I get the following error:

error: invalid operands of types 'double' and 'double' to binary 'operator%'

Here's the code:

int main() {
    double x = 6.3;
    double y = 2;
    double z = x % y;
}

Upvotes: 248

Views: 289094

Answers (5)

Pamputt
Pamputt

Reputation: 440

Here is a function that also works with both positive and negative numbers:

#include <cmath>

inline double nmod(double x, double y) {
    return std::fmod(std::fmod(x, y) + y, y);
}

It will always return a positive number if x is negative or positive.

Note that if y is negative, the returned value will be negative. If you need a function that will return a positive in any case, then one need to use std::fabs() on y, i.e.:

inline double nmod2(double x, double y) {
    return std::fmod(std::fmod(x, y) + std::fabs(y), y);
}

Upvotes: 1

Mysticial
Mysticial

Reputation: 471289

The % operator is for integers. You're looking for the fmod() function.

#include <cmath>

int main()
{
    double x = 6.3;
    double y = 2.0;
    double z = std::fmod(x,y);

}

Upvotes: 343

Mystical
Mystical

Reputation: 2783

You can implement your own modulus function to do that for you:

double dmod(double x, double y) {
    return x - (int)(x/y) * y;
}

Then you can simply use dmod(6.3, 2) to get the remainder, 0.3.

Upvotes: 10

Sceptical Jule
Sceptical Jule

Reputation: 917

Use fmod() from <cmath>. If you do not want to include the C header file:

template<typename T, typename U>
constexpr double dmod (T x, U mod)
{
    return !mod ? x : x - mod * static_cast<long long>(x / mod);
}

//Usage:
double z = dmod<double, unsigned int>(14.3, 4);
double z = dmod<long, float>(14, 4.6);
//This also works:
double z = dmod(14.7, 0.3);
double z = dmod(14.7, 0);
double z = dmod(0, 0.3f);
double z = dmod(myFirstVariable, someOtherVariable);

Upvotes: 6

MSN
MSN

Reputation: 54614

fmod(x, y) is the function you use.

Upvotes: 40

Related Questions