constexpr (C++)

Articol
02/22/2023

The keyword constexpr was introduced in C++11 and improved in C++14. It means constant expression. Like const, it can be applied to variables: A compiler error is raised when any code attempts to modify the value. Unlike const, constexpr can also be applied to functions and class constructors. constexpr indicates that the value, or return value, is constant and, where possible, is computed at compile time.

A constexpr integral value can be used wherever a const integer is required, such as in template arguments and array declarations. And when a value is computed at compile time instead of run time, it helps your program run faster and use less memory.

To limit the complexity of compile-time constant computations, and their potential impacts on compilation time, the C++14 standard requires the types in constant expressions to be literal types.

Syntax

constexpr literal-type identifier = constant-expression ;
constexpr literal-type identifier { constant-expression } ;
constexpr literal-type identifier ( params ) ;
constexpr ctor ( params ) ;

Parameters

params
One or more parameters, each of which must be a literal type and must itself be a constant expression.

Return value

A constexpr variable or function must return a literal type.

constexpr variables

The primary difference between const and constexpr variables is that the initialization of a const variable can be deferred until run time. A constexpr variable must be initialized at compile time. All constexpr variables are const.

A variable can be declared with constexpr, when it has a literal type and is initialized. If the initialization is performed by a constructor, the constructor must be declared as constexpr.
A reference may be declared as constexpr when both these conditions are met: The referenced object is initialized by a constant expression, and any implicit conversions invoked during initialization are also constant expressions.
All declarations of a constexpr variable or function must have the constexpr specifier.

constexpr float x = 42.0;
constexpr float y{108};
constexpr float z = exp(5, 3);
constexpr int i; // Error! Not initialized
int j = 0;
constexpr int k = j + 1; //Error! j not a constant expression

constexpr functions

A constexpr function is one whose return value is computable at compile time when consuming code requires it. Consuming code requires the return value at compile time to initialize a constexpr variable, or to provide a non-type template argument. When its arguments are constexpr values, a constexpr function produces a compile-time constant. When called with non-constexpr arguments, or when its value isn't required at compile time, it produces a value at run time like a regular function. (This dual behavior saves you from having to write constexpr and non-constexpr versions of the same function.)

A constexpr function or constructor is implicitly inline.

The following rules apply to constexpr functions:

A constexpr function must accept and return only literal types.
A constexpr function can be recursive.
Before C++20, a constexpr function can't be virtual, and a constructor can't be defined as constexpr when the enclosing class has any virtual base classes. In C++20 and later, a constexpr function can be virtual. Visual Studio 2019 version 16.10 and later versions support constexpr virtual functions when you specify the /std:c++20 or later compiler option.
The body can be defined as = default or = delete.
The body can contain no goto statements or try blocks.
An explicit specialization of a non-constexpr template can be declared as constexpr:
An explicit specialization of a constexpr template doesn't also have to be constexpr:

The following rules apply to constexpr functions in Visual Studio 2017 and later:

It may contain if and switch statements, and all looping statements including for, range-based for, while, and do-while.
It may contain local variable declarations, but the variable must be initialized. It must be a literal type, and can't be static or thread-local. The locally declared variable isn't required to be const, and may mutate.
A constexpr non-static member function isn't required to be implicitly const.

constexpr float exp(float x, int n)
{
    return n == 0 ? 1 :
        n % 2 == 0 ? exp(x * x, n / 2) :
        exp(x * x, (n - 1) / 2) * x;
}

Tip

In the Visual Studio debugger, when debugging a non-optimised Debug build, you can tell whether a constexpr function is being evaluated at compile time by putting a breakpoint inside it. If the breakpoint is hit, the function was called at run-time. If not, then the function was called at compile time.

extern constexpr

The /Zc:externConstexpr compiler option causes the compiler to apply external linkage to variables declared by using extern constexpr. In earlier versions of Visual Studio, either by default or when /Zc:externConstexpr- is specified, Visual Studio applies internal linkage to constexpr variables even when the extern keyword is used. The /Zc:externConstexpr option is available starting in Visual Studio 2017 Update 15.6, and is off by default. The /permissive- option doesn't enable /Zc:externConstexpr.

Example

The following example shows constexpr variables, functions, and a user-defined type. In the last statement in main(), the constexpr member function GetValue() is a run-time call because the value isn't required to be known at compile time.

// constexpr.cpp
// Compile with: cl /EHsc /W4 constexpr.cpp
#include <iostream>

using namespace std;

// Pass by value
constexpr float exp(float x, int n)
{
    return n == 0 ? 1 :
        n % 2 == 0 ? exp(x * x, n / 2) :
        exp(x * x, (n - 1) / 2) * x;
}

// Pass by reference
constexpr float exp2(const float& x, const int& n)
{
    return n == 0 ? 1 :
        n % 2 == 0 ? exp2(x * x, n / 2) :
        exp2(x * x, (n - 1) / 2) * x;
}

// Compile-time computation of array length
template<typename T, int N>
constexpr int length(const T(&)[N])
{
    return N;
}

// Recursive constexpr function
constexpr int fac(int n)
{
    return n == 1 ? 1 : n * fac(n - 1);
}

// User-defined type
class Foo
{
public:
    constexpr explicit Foo(int i) : _i(i) {}
    constexpr int GetValue() const
    {
        return _i;
    }
private:
    int _i;
};

int main()
{
    // foo is const:
    constexpr Foo foo(5);
    // foo = Foo(6); //Error!

    // Compile time:
    constexpr float x = exp(5, 3);
    constexpr float y { exp(2, 5) };
    constexpr int val = foo.GetValue();
    constexpr int f5 = fac(5);
    const int nums[] { 1, 2, 3, 4 };
    const int nums2[length(nums) * 2] { 1, 2, 3, 4, 5, 6, 7, 8 };

    // Run time:
    cout << "The value of foo is " << foo.GetValue() << endl;
}

Requirements

Visual Studio 2015 or later.

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