int类型的无效操作数和二进制'operator％'的double

Question

After compiling the program I am getting below error

invalid operands of types int and double to binary 'operator%' at line 
"newnum1 = two % (double)10.0;"

Why is it so?

#include<iostream>
#include<math>
using namespace std;
int main()
{
    int num;
    double two = 1;
    double newnum, newnum1;
    newnum = newnum1 = 0;
    for(num = 1; num <= 50; num++)
    {

        two = two * 2;
    }
    newnum1 = two % (double)10.0;
    newnum = newnum + newnum1;
    cout << two << "\n";
    return 0;
}

Answer 1

Because % is only defined for integer types. That's the modulus operator.

5.6.2 of the standard:

The operands of * and / shall have arithmetic or enumeration type; the operands of % shall have integral or enumeration type. [...]

As Oli pointed out, you can use fmod() . Don't forget to include math.h .

Answer 2

Because % only works with integer types. Perhaps you want to use fmod() .

Answer 3

Yes. % operator is not defined for double type. Same is true for bitwise operators like "&,^,|,~,<<,>>" as well.