The Modulus Operator (%)

We've previously seen that modular division is the default form used in C++ when working with integers.

The / operator lets us get the quotient:

1// This will be 1
2int Quotient { 5 / 3 };

To get the remainder, we use the modulus operator, denoted by %:

1// This will be 2
2int Remainder { 5 % 3 };

A common use of the modulus operator is to check if a number is even or odd. After being divided by $2$, an even number will have a remainder of $0$, whilst an odd number will have a remainder of $1$:

1bool isEven(int Number) {
2  return Number % 2 == 0;
3}

1int Remainder { 5 % 6 };

One of the useful properties of modular arithmetic is its tendency to "wrap around" after they reach a certain value.

Let's see an example:

10 % 3; // 0
21 % 3; // 1
32 % 3; // 2
4
53 % 3; // 0
64 % 3; // 1
75 % 3; // 2
8
96 % 3; // 0
107 % 3; // 1
118 % 3; // 2

We can see a repeating pattern developing here. As we increment the left operand, the modulus operator returns a repeating cycle of $0, 1, 2$. The length of this cycle is determined by the right operand. In this case, our right operand was $3$, so we got a repeating pattern of 3 numbers.

If our right operand were $2$ we'd get a repeating cycle of $0$ and $1$. Were it $10$, our cycle would be $0, 1, ..., 9$

This repeating pattern is one of the main reasons that the modulus is useful in programming - particularly when paired with loops.

The "wrapping around" nature of the modulus operator gives us an easy way to generate a boolean expression that is true on some subset of our loop iterations.

For example, if we increment a variable i on every iteration of our loop, i % 3 will be 0 on every third iteration.

Below, we use this calculation within an if statement to insert a line break on every third iteration:

1#include <iostream>
2using namespace std;
3
4int main(){
5  for (int i = 1; i <= 9; ++i) {
6    cout << i << " ";
7    if (i % 3 == 0) cout << '\n'; 
8  }
9}

11 2 3 
24 5 6 
37 8 9

The next lesson introduces a famous programming challenge where the modulus operator is crucial to solving the problem.