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

[jktruimp1]

I am trying to reduce fractions.
If I replace double by int,there is no error,and it will run answers. But the wrong one.....
I don't know if I use the reducing fractions' rule wrongly.
Can someone help me pleaseQQ

RationalNumber::RationalNumber(double a,double b)
{
    if(b>0)//prvent deno<0
    {
        double r;
        if(a<0)//if nume<0
        {
            a=-a;
            r=(a%b);
            while(r!=0)
            {
                a=b;
                b=r;
                r=a%b;
            }
            this->nume=-a;
            this->deno=b;
            cout<<"test2"<<nume<<"/"<<deno<<endl;
        }
        else
        {
            r=a%b;
            while(r!=0)
            {
                a=b;
                b=r;
                r=a%b;
            }
 
            this->nume=a;
            this->deno=b;
            cout<<"test2"<<nume<<"/"<<deno<<endl;
        }
    }
    else
    {
        cout<<"It's a invalid fraction."<<endl;
        cout<<"The fraction will be set to 0/1"<<endl;
        this->nume=0;
        this->deno=1;
        cout<<"test2"<<nume<<"/"<<deno<<endl;
    }
}
 
 

[weaknessforcats]

The modulus operator % applies only to integers.

In set theory, the modulo 5 set contains only 0,1,2,3,4. So if you have an element of another set, say the base 10 integer number system, like 17, you convert to modulo five by 17 % 5 which produces 2. That 2 is the element value in the modulo 5 set.

Floating point, on the other hand, has no conversion to modulo 5 because the values of floating point are not defined. That is, 123.456..... has decimal places that go on forever. It can't be pinned down to a fixed value. Hence, no modulus conversion.

[donbock]

From a mathematics point of view, a rational number is defined as the ratio of two integers. I don't know what you get from the ratio of two real numbers.

[Zacariaz]

it should also be noted that those =-a statements could potentially, though not likely, cause trouble. =(-a) would probably better, or at the very least some spacing.