5.3 — Modulus and Exponentiation

The modulus operator

The modulus operator (also informally known as the remainder operator) is an operator that returns the remainder after doing an integer division. For example, 7 / 4 = 1 remainder 3. Therefore, 7 % 4 = 3. As another example, 25 / 7 = 3 remainder 4, thus 25 % 7 = 4. Modulus only works with integer operands.

Modulus is most useful for testing whether a number is evenly divisible by another number: if x % y evaluates to 0, then we know that x is evenly divisible by y.

Here are a couple runs of this program:

Enter an integer: 6
Enter another integer: 3
The remainder is: 0
6 is evenly divisible by 3
Enter an integer: 6
Enter another integer: 4
The remainder is: 2
6 is not evenly divisible by 4

Modulus with negative numbers

The modulus operator can also work with negative operands. x % y always returns results with the sign of x.

Running the above program:

Enter an integer: -6
Enter another integer: 4
The remainder is: -2
-6 is not evenly divisible by 4
Enter an integer: 6
Enter another integer: -4
The remainder is: 2
6 is not evenly divisible by -4

In both cases, you can see the remainder takes the sign of the first operand.


Prior to C++11, modulus with a negative operand could result in either a positive or negative result. This was made predictable in C++11.

Where’s the exponent operator?

You’ll note that the ^ operator (commonly used to denote exponentiation in mathematics) is a Bitwise XOR operation in C++ (covered in lesson O.3 -- Bit manipulation with bitwise operators and bit masks). C++ does not include an exponent operator.

To do exponents in C++, #include the <cmath> header, and use the pow() function:

Note that the parameters (and return value) of function pow() are of type double. Due to rounding errors in floating point numbers, the results of pow() may not be precise (even if you pass it integers or whole numbers).

If you want to do integer exponentiation, you’re best off using your own function to do so. The following function implements integer exponentiation (using the non-intuitive “exponentiation by squaring” algorithm for efficiency):

Don’t worry if you don’t understand how this function works -- you don’t need to understand it in order to call it.




In the vast majority of cases, integer exponentiation will overflow the integral type. This is likely why such a function wasn’t included in the standard library in the first place.

Quiz time

Question #1

What does the following expression evaluate to? 6 + 5 * 4 % 3

Show Solution

Question #2

Write a program that asks the user to input an integer, and tells the user whether the number is even or odd. Write a function called isEven() that returns true if an integer passed to it is even, and false otherwise. Use the modulus operator to test whether the integer parameter is even.

Hint: You’ll need to use if statements and the comparison operator (==) for this program. See lesson 4.9 -- Boolean values if you need a refresher on how to do this.

Your program should match the following output:

Enter an integer: 5
5 is odd

Show Solution

5.4 -- Increment/decrement operators, and side effects
5.2 -- Arithmetic operators

73 comments to 5.3 — Modulus and Exponentiation

  • Instead of using the % operator I would like to be able to read the first bit of a variable
    If the first bit of the variable is 1, the value will be odd if it is 0 it will be pair.

    00000111 = 7 odd
    00001101 = 13 odd
    00000100 = 4 pair
    10000000 = 128 pair
    00000001 = 1 odd
    00000010 = 2 pair

    Could someone help me with the code, I don't know how to make it

    • nascardriver

  • evolve

    will this code work same?

  • John

    Here's another interesting way to solve the question by exploiting the fact that division of integers drops the fractional part.

  • McKeon

    You were right about how i was tempted to write the code shown in Question #2.

    Thats been my main concern about this whole learning experience is that i know how to utilize what ive learned and get it to work, but learning other ways to do it that are either more effecient and or "more or less complex", Are my main concern.

    I need to get better at the problem solving of things and the logic of the writing the programs. I'm sure i will get better as i go but do you have a tip for me to learn better ways for me to learn how to better problem solve about how i need to setup what im about to write. its hard for me to word but i think you know what i am trying to say.

    It's just like how when you learn math in school you are like i will never use this but you dont realize you actually will use it all the time because it improves your problem solving skills and the way you think about things it rewires your brain. That is what i want to get better at is coming up with better ways. maybe i just answered my own question.

    But how you had shown "you may have been tempted to write it this way" and i did and sure it worked. But then you shown "well you can do this instead you dont have to itterate so much and there is an easier way to do this" thats what inspires me with coding and what i find rewarding is when i can do these little excercises and they look exactly like your solution and show me another better way i could have done it.

  • Andreas Krug

    Hello Alex & nascardriver,

    in chapter 2.11 - Header files (The #include order of header files) you wrote to sort the headers alphabetically, but the third code example from (Where’s the exponent operator?) in this chapter has iostream before cstdint.

    Keep up the great work,

  • It's+Me

    bool IsEven(int x) {
        if (x % 2 == 0) {
            std::cout << x <<" is even to ";
        else {
            std::cout << x << " is odd to ";
        return false;
    int main() {
        std::cout << "Enter an integer: ";
        int x;
        std::cin >> x;
        std::cout << IsEven(x);


    // not so perfect as your program though :(

  • Amir

    hope this is okay

    • Anders Nesse

      Hi Amir.

      My problem with this function is that it's not intuitive for other users. The function name is "isEven", but returns 0 (false) if the number is even, and 1 (true) if the number is odd. A better name for the function would be isOdd(), but since the quiz asks for a function by the name isEven(), my solution would be as follows:

      This code does what the name suggests, i.e. to evaluate whether or not the entered number is even (true/false), by using the ! (logical not) operator. It can still be a bit of a hassle to understand intuitively from an outsider's perspective by just looking at the code, which is (I assume) why Alex and Nascardriver chose their solution instead:

      • Amir

        you right :)  it makes sense . I wasn't caring that much about how to make code readable and easy to understand. Now after learning more I really see how its  important to make each statement clear . thank you :)

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