12.2 — Value categories (lvalues and rvalues)

Before we talk about our first compound type (lvalue references), we’re going to take a little detour and talk about what an lvalue is.

In lesson 1.10 -- Introduction to expressions, we defined an expression as “a combination of literals, variables, operators, and function calls that can be executed to produce a singular value”.

For example:

#include <iostream>

int main()
{
    std::cout << 2 + 3 << '\n'; // The expression 2 + 3 produces the value 5

    return 0;
}

In the above program, the expression 2 + 3 is evaluated to produce the value 5, which is then printed to the console.

In lesson 6.4 -- Increment/decrement operators, and side effects, we also noted that expressions can produce side effects that outlive the expression:

#include <iostream>

int main()
{
    int x { 5 };
    ++x; // This expression statement has the side-effect of incrementing x
    std::cout << x << '\n'; // prints 6

    return 0;
}

In the above program, the expression ++x increments the value of x, and that value remains changed even after the expression has finished evaluating.

Besides producing values and side effects, expressions can do one more thing: they can evaluate to objects or functions. We’ll explore this point further in just a moment.

The properties of an expression

To help determine how expressions should evaluate and where they can be used, all expressions in C++ have two properties: a type and a value category.

The type of an expression

The type of an expression is equivalent to the type of the value, object, or function that results from the evaluated expression. For example:

int main()
{
    auto v1 { 12 / 4 }; // int / int => int
    auto v2 { 12.0 / 4 }; // double / int => double

    return 0;
}

For v1, the compiler will determine (at compile time) that a division with two int operands will produce an int result, so int is the type of this expression. Via type inference, int will then be used as the type of v1.

For v2, the compiler will determine (at compile time) that a division with a double operand and an int operand will produce a double result. Remember that arithmetic operators must have operands of matching types, so in this case, the int operand gets converted to a double, and a floating point division is performed. So double is the type of this expression.

The compiler can use the type of an expression to determine whether an expression is valid in a given context. For example:

#include <iostream>

void print(int x)
{
    std::cout << x << '\n';
}

int main()
{
    print("foo"); // error: print() was expecting an int argument, we tried to pass in a string literal

    return 0;
}

In the above program, the print(int) function is expecting an int parameter. However, the type of the expression we’re passing in (the string literal "foo") does not match, and no conversion can be found. So a compile error results.

Note that the type of an expression must be determinable at compile time (otherwise type checking and type deduction wouldn’t work) -- however, the value of an expression may be determined at either compile time (if the expression is constexpr) or runtime (if the expression is not constexpr).

The value category of an expression

Now consider the following program:

int main()
{
    int x{};

    x = 5; // valid: we can assign 5 to x
    5 = x; // error: can not assign value of x to literal value 5

    return 0;
}

One of these assignment statements is valid (assigning value 5 to variable x) and one is not (what would it mean to assign the value of x to the literal value 5?). So how does the compiler know which expressions can legally appear on either side of an assignment statement?

The answer lies in the second property of expressions: the value category. The value category of an expression (or subexpression) indicates whether an expression resolves to a value, a function, or an object of some kind.

Prior to C++11, there were only two possible value categories: lvalue and rvalue.

In C++11, three additional value categories (glvalue, prvalue, and xvalue) were added to support a new feature called move semantics.

Lvalue and rvalue expressions

An lvalue (pronounced “ell-value”, short for “left value” or “locator value”, and sometimes written as “l-value”) is an expression that evaluates to an identifiable object or function (or bit-field).

The term “identity” is used by the C++ standard, but is not well-defined. An entity (such as an object or function) that has an identity can be differentiated from other similar entities (typically by comparing the addresses of the entity).

Entities with identities can be accessed via an identifier, reference, or pointer, and typically have a lifetime longer than a single expression or statement.

int main()
{
    int x { 5 };
    int y { x }; // x is an lvalue expression

    return 0;
}

In the above program, the expression x is an lvalue expression as it evaluates to variable x (which has an identifier).

Since the introduction of constants into the language, lvalues come in two subtypes: a modifiable lvalue is an lvalue whose value can be modified. A non-modifiable lvalue is an lvalue whose value can’t be modified (because the lvalue is const or constexpr).

int main()
{
    int x{};
    const double d{};

    int y { x }; // x is a modifiable lvalue expression
    const double e { d }; // d is a non-modifiable lvalue expression

    return 0;
}

An rvalue (pronounced “arr-value”, short for “right value”, and sometimes written as r-value) is an expression that is not an lvalue. Rvalue expressions evaluate to a value. Commonly seen rvalues include literals (except C-style string literals, which are lvalues) and the return value of functions and operators that return by value. Rvalues aren’t identifiable (meaning they have to be used immediately), and only exist within the scope of the expression in which they are used.

int return5()
{
    return 5;
}

int main()
{
    int x{ 5 }; // 5 is an rvalue expression
    const double d{ 1.2 }; // 1.2 is an rvalue expression

    int y { x }; // x is a modifiable lvalue expression
    const double e { d }; // d is a non-modifiable lvalue expression
    int z { return5() }; // return5() is an rvalue expression (since the result is returned by value)

    int w { x + 1 }; // x + 1 is an rvalue expression
    int q { static_cast<int>(d) }; // the result of static casting d to an int is an rvalue expression

    return 0;
}

You may be wondering why return5(), x + 1, and static_cast<int>(d) are rvalues: the answer is because these expressions produce temporary values that are not identifiable objects.

Value categories and operators

Unless otherwise specified, operators expect their operands to be rvalues. For example, binary operator+ expects its operands to be rvalues:

#include <iostream>

int main()
{
    std::cout << 1 + 2; // 1 and 2 are rvalues, operator+ returns an rvalue

    return 0;
}

The literals 1 and 2 are both rvalue expressions. operator+ will happily use these to return the rvalue expression 3.

Now we can answer the question about why x = 5 is valid but 5 = x is not: an assignment operation requires its left operand to be a modifiable lvalue expression. The latter assignment (5 = x) fails because the left operand expression 5 is an rvalue, not a modifiable lvalue.

int main()
{
    int x{};

    // Assignment requires the left operand to be a modifiable lvalue expression and the right operand to be an rvalue expression
    x = 5; // valid: x is a modifiable lvalue expression and 5 is an rvalue expression
    5 = x; // error: 5 is an rvalue expression and x is a modifiable lvalue expression

    return 0;
}

Lvalue-to-rvalue conversion

Since assignment operations expect the right operand to be an rvalue expression, you might be wondering why the following works:

int main()
{
    int x{ 1 };
    int y{ 2 };

    x = y; // y is not an rvalue, but this is legal

    return 0;
}

In cases where an rvalue is expected but an lvalue is provided, the lvalue will undergo an lvalue-to-rvalue conversion so that it can be used in such contexts. This basically means the lvalue is evaluated to produce its value, which is an rvalue.

In the above example, the lvalue expression y undergoes an lvalue-to-rvalue conversion, which evaluates y to produce an rvalue (2), which is then assigned to x.

Now consider this example:

int main()
{
    int x { 2 };

    x = x + 1;

    return 0;
}

In this statement, the variable x is being used in two different contexts. On the left side of the assignment operator (where an lvalue expression is required), x is an lvalue expression that evaluates to variable x. On the right side of the assignment operator, x undergoes an lvalue-to-rvalue conversion and is then evaluated so that its (2) can be used as the left operand of operator+. operator+ returns the rvalue expression 3, which is then used as the right operand for the assignment.

How to differentiate lvalues and rvalues

You may still be confused about what kind of expressions qualify as an lvalue vs an rvalue. For example, is the result of operator++ an lvalue or an rvalue? We’ll cover various methods you can use to determine which is which here.

For a more complete list of lvalue and rvalue expressions, you can consult technical documentation.

Finally, we can write a program and have the compiler tell us what kind of expression something is. The following code demonstrates a method that determines whether an expression is an lvalue or an rvalue:

#include <iostream>
#include <string>

// T& is an lvalue reference, so this overload will be preferred for lvalues
template <typename T>
constexpr bool is_lvalue(T&)
{
    return true;
}

// T&& is an rvalue reference, so this overload will be preferred for rvalues
template <typename T>
constexpr bool is_lvalue(T&&)
{
    return false;
}

// A helper macro (#expr prints whatever is passed in for expr as text)
#define PRINTVCAT(expr) { std::cout << #expr << " is an " << (is_lvalue(expr) ? "lvalue\n" : "rvalue\n"); }

int getint() { return 5; }

int main()
{
    PRINTVCAT(5);        // rvalue
    PRINTVCAT(getint()); // rvalue
    int x { 5 };
    PRINTVCAT(x);        // lvalue
    PRINTVCAT(std::string {"Hello"}); // rvalue
    PRINTVCAT("Hello");  // lvalue
    PRINTVCAT(++x);      // lvalue
    PRINTVCAT(x++);      // rvalue
} 

This prints:

5 is an rvalue
getint() is an rvalue
x is an lvalue
std::string {"Hello"} is an rvalue
"Hello" is an lvalue
++x is an lvalue
x++ is an rvalue

This method relies on two overloaded functions: one with an lvalue refrence parameter and one with an rvalue reference parameter. The lvalue reference version will be preferred for lvalue arguments, and the rvalue reference version will be preferred for rvalue arguments. Thus we can determine whether the argument is an lvalue or rvalue based on which function gets selected.

So as you can see, whether operator++ results in an lvalue or an rvalue depends on whether it is used as a prefix operator (which returns an lvalue) or a postfix operator (which returns an rvalue)!

Now that we’ve covered lvalues, we can get to our first compound type: the lvalue reference.