10.5 — Arithmetic conversions

In lesson 6.1 -- Operator precedence and associativity, we discussed how expressions are evaluated according to the precedence and associativity of their operators.

Consider the following expression:

int x { 2 + 3 };

Binary operator+ is given two operands, both of type int. Because both operands are of the same type, that type will be used to perform the calculation, and the value returned will also be of this same type. Thus, 2 + 3 will evaluate to int value 5.

But what happens when the operands of a binary operator are of different types?

??? y { 2 + 3.5 };

In this case, operator+ is being given one operand of type int and another of type double. Should the result of the operator be returned as an int, a double, or possibly something else altogether?

In C++, certain operators require that their operands be of the same type. If one of these operators is invoked with operands of different types, one or both of the operands will be implicitly converted to matching types using a set of rules called the usual arithmetic conversions. The matching type produced as a result of the usual arithmetic conversion rules is called the common type of the operands.

The operators that require operands of the same type

The following operators require their operands to be of the same type:

• The binary arithmetic operators: +, -, *, /, %
• The binary relational operators: <, >, <=, >=, ==, !=
• The binary bitwise arithmetic operators: &, ^, |
• The conditional operator ?: (excluding the condition, which is expected to be of type bool)

The usual arithmetic conversion rules

The usual arithmetic conversion rules are somewhat complex, so we’ll simplify a bit. The compiler has a ranked list of types that looks something like this:

• long double (highest rank)
• double
• float
• long long
• long
• int (lowest rank)

The following rules are applied to find a matching type:

Step 1:

• If one operand is an integral type and the other a floating point type, the integral operand is converted to the type of the floating point operand (no integral promotion takes place).
• Otherwise, any integral operands are numerically promoted (see 10.2 -- Floating-point and integral promotion).

Step 2:

• After promotion, if one operand is signed and the other unsigned, special rules apply (see below)
• Otherwise, the operand with lower rank is converted to the type of the operand with higher rank.

Some examples

In the following examples, we’ll use the typeid operator (included in the <typeinfo> header), to show the resulting type of an expression.

First, let’s add an int and a double:

#include <iostream>
#include <typeinfo> // for typeid()

int main()
{
    int i{ 2 };
    std::cout << typeid(i).name() << '\n'; // show us the name of the type for i

    double d{ 3.5 };
    std::cout << typeid(d).name() << '\n'; // show us the name of the type for d

    std::cout << typeid(i + d).name() << ' ' << i + d << '\n'; // show us the type of i + d

    return 0;
}

In this case, the double operand has the highest priority, so the lower priority operand (of type int) is type converted to double value 2.0. Then double values 2.0 and 3.5 are added to produce double result 5.5.

On the author’s machine, this prints:

int
double
double 5.5

Note that your compiler may display something slightly different, as the names output by typeid.name() are implementation-specific.

Now let’s add two values of type short:

#include <iostream>
#include <typeinfo> // for typeid()

int main()
{
    short a{ 4 };
    short b{ 5 };
    std::cout << typeid(a + b).name() << ' ' << a + b << '\n'; // show us the type of a + b

    return 0;
}

Because neither operand appears on the priority list, both operands undergo integral promotion to type int. The result of adding two ints is an int, as you would expect:

int 9

Signed and unsigned issues

This prioritization hierarchy and conversion rules can cause some problematic issues when mixing signed and unsigned values. For example, take a look at the following code:

#include <iostream>
#include <typeinfo> // for typeid()

int main()
{
    std::cout << typeid(5u-10).name() << ' ' << 5u - 10 << '\n'; // 5u means treat 5 as an unsigned integer

    return 0;
}

You might expect the expression 5u - 10 to evaluate to -5 since 5 - 10 = -5. But here’s what actually results:

unsigned int 4294967291

Due to the conversion rules, the int operand is converted to an unsigned int. And since the value -5 is out of range of an unsigned int, we get a result we don’t expect.

Here’s another example showing a counterintuitive result:

#include <iostream>

int main()
{
    std::cout << std::boolalpha << (-3 < 5u) << '\n';

    return 0;
}

While it’s clear to us that 5 is greater than -3, when this expression evaluates, -3 is converted to a large unsigned int that is larger than 5. Thus, the above prints false rather than the expected result of true.

This is one of the primary reasons to avoid unsigned integers -- when you mix them with signed integers in arithmetic expressions, you’re at risk for unexpected results. And the compiler probably won’t even issue a warning.

std::common_type and std::common_type_t

In future lessons, we’ll encounter cases where it is useful to know what the common type of two type is. std::common_type and the useful type alias std::common_type_t (both defined in the <type_traits> header) can be used for just this purpose.

For example, std::common_type_t<int, double> returns the common type of int and double, and std::common_type_t<unsigned int, long> returns the common type of unsigned int and long.

We’ll show an example where this is useful in lesson 11.8 -- Function templates with multiple template types.