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constexpr specifier (since C++11) - cppreference.com

• constexpr - specifies that the value of a variable, structured binding(since C++26) or function can appear in constant expressions

The constexpr specifier declares that it is possible to evaluate the value of the entities at compile time. Such entities can then be used where only compile time constant expressions are allowed (provided that appropriate function arguments are given).

A constexpr specifier used in an object declaration or non-static member function(until C++14) implies const.

A constexpr specifier used in the first declaration of a function or static data member(since C++17) implies inline. If any declaration of a function or function template has a constexpr specifier, then every declaration must contain that specifier.

A variable or variable template(since C++14) can be declared constexpr if all following conditions are satisfied:

• The declaration is a definition.
• It is of a literal type.
• It is initialized (by the declaration).

• It has constant destruction, which means one of the following conditions needs to be satisfied:

• It is not of class type nor (possibly multi-dimensional) array thereof.
• It is of a class type with a constexpr destructor or (possibly multi-dimensional) array thereof, and for a hypothetical expression e whose only effect is to destroy the object, e would be a core constant expression if the lifetime of the object and its non-mutable subobjects (but not its mutable subobjects) were considered to start within e.

If a constexpr variable is not translation-unit-local, it should not be initialized to refer to a translation-unit-local entity that is usable in constant expressions, nor have a subobject that refers to such an entity. Such initialization is disallowed in a module interface unit (outside its private module fragment, if any) or a module partition, and is deprecated in any other context.

A function or function template can be declared constexpr.

A function is constexpr-suitable if all following conditions are satisfied:

• If it is a constructor or destructor(since C++20), its class does not have any virtual base class.

• It is not a virtual function.

• Its return type (if exists) is a literal type.
• Each of its parameter types is a literal type.

• It is not a coroutine.

• Its function body is = default, = delete, or a compound statement enclosing only the following:

• null statements
• static_assert declarations
• typedef declarations and alias declarations that do not define classes or enumerations
• using declarations
• using directives
• exactly one return statement if the function is not a constructor

• Its function body is = default, = delete, or a compound statement that(until C++20) does not enclose the following:

• goto statements
• statements with labels other than case and default

• definitions of variables of non-literal types
• definitions of variables of static or thread storage duration

Except for instantiated constexpr functions, non-templated constexpr functions must be constexpr-suitable.

For a non-constructor constexpr function that is neither defaulted nor templated, if no argument values exist such that an invocation of the function could be an evaluated subexpression of a core constant expression, the program is ill-formed, no diagnostic required.

For a templated constexpr function, if no specialization of the function/class template would make the templated function constexpr-suitable when considered as a non-templated function, the program is ill-formed, no diagnostic required.

An invocation of a constexpr function in a given context produces the same result as an invocation of an equivalent non-constexpr function in the same context in all respects, with the following exceptions:

• An invocation of a constexpr function can appear in a constant expression.
• Copy elision is not performed in a constant expression.

On top of the requirements of constexpr functions, a constructor also needs to satisfy all following conditions to be constexpr-suitable:

• Its function body is = delete or satisfies the following additional requirements:

• If the class is a union having variant members, exactly one of them is initialized.
• If the class is a union-like class, but is not a union, for each of its anonymous union members having variant members, exactly one of them is initialized.
• Every non-variant non-static data member and base class subobject is initialized.

• If the constructor is a delegating constructor, the target constructor is a constexpr constructor.
• If the constructor is a non-delegating constructor, every constructor selected to initialize non-static data members and base class subobjects is a constexpr constructor.

• The class does not have any virtual base class.

For a constexpr constructor that is neither defaulted nor templated, if no argument values exist such that an invocation of the function could be an evaluated subexpression of the initialization full-expression of some object subject to constant expression, the program is ill-formed, no diagnostic required.

Destructors cannot be constexpr, but a trivial destructor can be implicitly called in constant expressions.

On top of the requirements of constexpr functions, a destructor also needs to satisfy all following conditions to be constexpr-suitable:

• For every subobject of class type or (possibly multi-dimensional) array thereof, that class type has a constexpr destructor.

• The class does not have any virtual base class.

Because the noexcept operator always returns true for a constant expression, it can be used to check if a particular invocation of a constexpr function takes the constant expression branch:

constexpr int f(); 
constexpr bool b1 = noexcept(f()); // false, undefined constexpr function
constexpr int f() { return 0; }
constexpr bool b2 = noexcept(f()); // true, f() is a constant expression

It is possible to write a constexpr function whose invocation can never satisfy the requirements of a core constant expression:

void f(int& i) // not a constexpr function
{
    i = 0;
}
 
constexpr void g(int& i) // well-formed since C++23
{
    f(i); // unconditionally calls f, cannot be a constant expression
}

Constexpr constructors are permitted for classes that are not literal types. For example, the default constructor of std::shared_ptr is constexpr, allowing constant initialization.

Reference variables can be declared constexpr (their initializers have to be reference constant expressions):

static constexpr int const& x = 42; // constexpr reference to a const int object
                                    // (the object has static storage duration
                                    //  due to life extension by a static reference)

Even though try blocks and inline assembly are allowed in constexpr functions, throwing exceptions that are uncaught(since C++26) or executing the assembly is still disallowed in a constant expression.

If a variable has constant destruction, there is no need to generate machine code in order to call destructor for it, even if its destructor is not trivial.

A non-lambda, non-special-member, and non-templated constexpr function cannot implicitly become an immediate function. Users need to explicitly mark it consteval to make such an intended function definition well-formed.

constexpr

Defines C++11/14 constexpr functions that compute factorials; defines a literal type that extends string literals:

#include <iostream>
#include <stdexcept>
 
// C++11 constexpr functions use recursion rather than iteration
constexpr int factorial(int n)
{
    return n <= 1 ? 1 : (n * factorial(n - 1));
}
 
// C++14 constexpr functions may use local variables and loops
#if __cplusplus >= 201402L
constexpr int factorial_cxx14(int n)
{
    int res = 1;
    while (n > 1)
        res *= n--;
    return res;
}
#endif // C++14
 
// A literal class
class conststr
{
    const char* p;
    std::size_t sz;
public:
    template<std::size_t N>
    constexpr conststr(const char(&a)[N]): p(a), sz(N - 1) {}
 
    // constexpr functions signal errors by throwing exceptions
    // in C++11, they must do so from the conditional operator ?:
    constexpr char operator[](std::size_t n) const
    {
        return n < sz ? p[n] : throw std::out_of_range("");
    }
 
    constexpr std::size_t size() const { return sz; }
};
 
// C++11 constexpr functions had to put everything in a single return statement
// (C++14 does not have that requirement)
constexpr std::size_t countlower(conststr s, std::size_t n = 0,
                                             std::size_t c = 0)
{
    return n == s.size() ? c :
        'a' <= s[n] && s[n] <= 'z' ? countlower(s, n + 1, c + 1)
                                   : countlower(s, n + 1, c);
}
 
// An output function that requires a compile-time constant, for testing
template<int n>
struct constN
{
    constN() { std::cout << n << '\n'; }
};
 
int main()
{
    std::cout << "4! = ";
    constN<factorial(4)> out1; // computed at compile time
 
    volatile int k = 8; // disallow optimization using volatile
    std::cout << k << "! = " << factorial(k) << '\n'; // computed at run time
 
    std::cout << "The number of lowercase letters in \"Hello, world!\" is ";
    constN<countlower("Hello, world!")> out2; // implicitly converted to conststr
 
    constexpr int a[12] = {0, 1, 2, 3, 4, 5, 6, 7, 8};
    constexpr int length_a = sizeof a / sizeof(int); // std::size(a) in C++17,
                                                      // std::ssize(a) in C++20
    std::cout << "Array of length " << length_a << " has elements: ";
    for (int i = 0; i < length_a; ++i)
        std::cout << a[i] << ' ';
    std::cout << '\n';
}

Output:

4! = 24
8! = 40320
The number of lowercase letters in "Hello, world!" is 9
Array of length 12 has elements: 0 1 2 3 4 5 6 7 8 0 0 0

The following behavior-changing defect reports were applied retroactively to previously published C++ standards.

1. ↑ It is redundant because there cannot be more than one declaration of a variable template with the constexpr specifier.

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