5.2 — Arithmetic operators

Unary arithmetic operators

There are two unary arithmetic operators, plus (+), and minus (-). As a reminder, unary operators are operators that only take one operand.

The unary minus operator returns the operand multiplied by -1. In other words, if x = 5, -x is -5.

The unary plus operator returns the value of the operand. In other words, +5 is 5, and +x is x. Generally you won’t need to use this operator since it’s redundant. It was added largely to provide symmetry with the unary minus operator.

For best effect, both of these operators should be placed immediately preceding the operand (e.g. -x, not - x).

Do not confuse the unary minus operator with the binary subtraction operator, which uses the same symbol. For example, in the expression x = 5 - -3;, the first minus is the binary subtraction operator, and the second is the unary minus operator.

Binary arithmetic operators

There are 5 binary arithmetic operators. Binary operators are operators that take a left and right operand.

The addition, subtraction, and multiplication operators work just like they do in real life, with no caveats.

Division and modulus (remainder) need some additional explanation. We’ll talk about division below, and modulus in the next lesson.

Integer and floating point division

It is easiest to think of the division operator as having two different “modes”.

If either (or both) of the operands are floating point values, the division operator performs floating point division. Floating point division returns a floating point value, and the fraction is kept. For example, 7.0 / 4 = 1.75, 7 / 4.0 = 1.75, and 7.0 / 4.0 = 1.75. As with all floating point arithmetic operations, rounding errors may occur.

If both of the operands are integers, the division operator performs integer division instead. Integer division drops any fractions and returns an integer value. For example, 7 / 4 = 1 because the fractional portion of the result is dropped. Similarly, -7 / 4 = -1 because the fraction is dropped.

Using static_cast<> to do floating point division with integers

The above raises the question -- if we have two integers, and want to divide them without losing the fraction, how would we do so?

In lesson 4.12 -- Introduction to type conversion and static_cast, we showed how we could use the static_cast<> operator to convert a char into an integer so it would print as an integer rather than a character.

We can similarly use static_cast<> to convert an integer to a floating point number so that we can do floating point division instead of integer division. Consider the following code:

#include <iostream>

int main()
{
    int x{ 7 };
    int y{ 4 };

    std::cout << "int / int = " << x / y << '\n';
    std::cout << "double / int = " << static_cast<double>(x) / y << '\n';
    std::cout << "int / double = " << x / static_cast<double>(y) << '\n';
    std::cout << "double / double = " << static_cast<double>(x) / static_cast<double>(y) << '\n';

    return 0;
}

This produces the result:

int / int = 1
double / int = 1.75
int / double = 1.75
double / double = 1.75

The above illustrates that if either operand is a floating point number, the result will be floating point division, not integer division.

Dividing by zero

Trying to divide by 0 (or 0.0) will generally cause your program to crash, as the results are mathematically undefined!

#include <iostream>

int main()
{
	std::cout << "Enter a divisor: ";
	int x{};
	std::cin >> x;

	std::cout << "12 / " << x << " = " << 12 / x << '\n';

	return 0;
}

If you run the above program and enter 0, your program will either crash or terminate abnormally. Go ahead and try it, it won’t harm your computer.

Arithmetic assignment operators

Up to this point, when you’ve needed to add 4 to a variable, you’ve likely done the following:

x = x + 4; // add 4 to existing value of x

This works, but it’s a little clunky, and takes two operators to execute (operator+, and operator=).

Because writing statements such as x = x + 4 is so common, C++ provides five arithmetic assignment operators for convenience. Instead of writing x = x + 4, you can write x += 4. Instead of x = x * y, you can write x *= y.

Thus, the above becomes:

x += 4; // add 4 to existing value of x