modulus is not working when using doubles in c++

Question

Im not being allowed to use mod. I think it's not working because I'm using doubles; is there a way around this? --- the commented areas arnt working

void displayResults(double num1, char oper, double num2)
{
     switch(oper)
     {
     case '+' :
     cout << num1 << "+" << num2 << "=" << (num1+num2) << endl;
     break;

     case '-' :
     cout << num1 << "-" << num2 << "=" << (num1-num2) << endl;
     break;

     case '*' :
     cout << num1 << "*" << num2 << "=" << (num1*num2) << endl;
     break; 

     case '/' :
          if ( num1==0 || num2==0)
          cout <<"A number divided by 0 or divided into 0 is always 0"<< endl;
          else
          cout << num1 << "/" << num2 << "=" << (num1/num2) /*+ (num1%num2) */ << endl;
          break; 
    // case '%' :
    // cout << num1 << "%" << num2 << "=" << (num1%num2);
    //break;
     }

}

You're not allowed? *How* is it not working? Compiler errors? Run-time Exceptions? Output not correct? — crashmstr, Oct 21 '13 at 19:32

score 5 · Answer 1 · answered Oct 21 '13 at 19:31

5

Use std::fmod. It has an overload for doubles:

#include <cmath>

std::fmod(num1, num2);

answered Oct 21 '13 at 19:31

David G

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I don't really get it. Mathematically, for example, what's the remainder of 2.6/1.5? – khajvah Oct 21 '13 at 19:34
1

@khajvah I may be Batman, but I'm no mathematician. :-) – David G Oct 21 '13 at 19:41
@khajvah: If we want the remainder of a/b, where b is positive, write a = q * b + r. q has to be an integer, and we need 0 <= r < b. r is the remainder. I think of b is negative, then in math we use a positive r, but in programming a negative r is often used. – Ken Wayne VanderLinde Oct 21 '13 at 19:45
@KenWayneVanderLinde According to `cppreference.com`, absolute value of `a-q*b` has to be smaller than absolute value of `b`, so when `b` is negative, we pick negative `q`. As a result, remainder is the same in both cases. – khajvah Oct 21 '13 at 20:09
@khajvah: Ah yes, fmod gives the same sign as `a`. But this differs from it mathematical usage, so it's quite confusing sometimes! – Ken Wayne VanderLinde Oct 21 '13 at 20:14

score 1 · Accepted Answer · answered Oct 21 '13 at 19:31

1

The modulus operator % is an integral function.

You need to use fmod for floating-point.

answered Oct 21 '13 at 19:31

John Dibling

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Shafik Yaghmour · Answer 3 · 2013-10-21T19:37:02.877

1

% is only allowed for integral types or unscoped enums, you can use std::fmod for double:

#include <cmath>
#include <iostream>

int main() {
    double num1 = 5.5;
    double num2 = 3.0;
    double z = std::fmod(num1,num2);

    std::cout << z << std::endl ;
    return 0;
}

edited Oct 21 '13 at 19:37

answered Oct 21 '13 at 19:31

Shafik Yaghmour

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score 0 · Answer 4 · answered Oct 21 '13 at 19:35

0

Of course, you could overload the '%' operator to perform the modulus operation on any type you like. There is no build-in support in the language for doubles, however.

answered Oct 21 '13 at 19:35

ThunderGr

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What you mean by "build-in". Isn't `std` "build-in"? There seems to be a function in std for doubles – khajvah Oct 21 '13 at 19:37
@khajvah there is a function, but not build in support for the *operator %* for doubles. – ThunderGr Oct 21 '13 at 19:37

score 0 · Answer 5 · answered Oct 21 '13 at 19:36

i think you can rather replace your doubles with 'long int' , that will help you overcome the error and this will also work with all integer based operations.

i made this change in one of my codes and got it right, with a warning though. But it did work just fine, so try that, it should work pretty well.

modulus is not working when using doubles in c++

5 Answers5