What does "default" mean after a class' function declaration?

Question

I've seen default used next to function declarations in a class. What does it do?

class C {
  C(const C&) = default;
  C(C&&) = default;
  C& operator=(const C&) & = default;
  C& operator=(C&&) & = default;
  virtual ~C() { }
};

user_895245 · Answer

C++17 N4659 standard draft

https://github.com/cplusplus/draft/blob/master/papers/n4659.pdf 11.4.2 "Explicitly-defaulted functions":

1 A function definition of the form:
attribute-specifier-seq opt decl-specifier-seq opt declarator virt-specifier-seq opt = default ;
is called an explicitly-defaulted definition. A function that is explicitly defaulted shall

(1.1) — be a special member function,

(1.2) — have the same declared function type (except for possibly differing ref-qualifiers and except that in the case of a copy constructor or copy assignment operator, the parameter type may be “reference to non-const T”, where T is the name of the member function’s class) as if it had been implicitly declared, and

(1.3) — not have default arguments.

2 An explicitly-defaulted function that is not defined as deleted may be declared constexpr only if it would have been implicitly declared as constexpr. If a function is explicitly defaulted on its first declaration, it is implicitly considered to be constexpr if the implicit declaration would be.

3 If a function that is explicitly defaulted is declared with a noexcept-specifier that does not produce the same exception specification as the implicit declaration (18.4), then

(3.1) — if the function is explicitly defaulted on its first declaration, it is defined as deleted;

(3.2) — otherwise, the program is ill-formed.

4 [ Example:
struct S {
  constexpr S() = default;            // ill-formed: implicit S() is not constexpr
  S(int a = 0) = default;             // ill-formed: default argument
  void operator=(const S&) = default; // ill-formed: non-matching return type
  ~ S() noexcept(false) = default;    // deleted: exception specification does not match
private:
  int i;                              // OK: private copy constructor
  S(S&);
};
S::S(S&) = default;                   // OK: defines copy constructor
— end example ]

5 Explicitly-defaulted functions and implicitly-declared functions are collectively called defaulted functions, and the implementation shall provide implicit definitions for them (15.1 15.4, 15.8), which might mean defining them as deleted. A function is user-provided if it is user-declared and not explicitly defaulted or deleted on its first declaration. A user-provided explicitly-defaulted function (i.e., explicitly defaulted after its first declaration) is defined at the point where it is explicitly defaulted; if such a function is implicitly defined as deleted, the program is ill-formed. [ Note: Declaring a function as defaulted after its first declaration can provide efficient execution and concise definition while enabling a stable binary interface to an evolving code base. — end note ]

6 [ Example:
struct trivial {
  trivial() = default;
  trivial(const trivial&) = default;
  trivial(trivial&&) = default;
  trivial& operator=(const trivial&) = default;
  trivial& operator=(trivial&&) = default;
  ~ trivial() = default;
};
struct nontrivial1 {
  nontrivial1();
};
nontrivial1::nontrivial1() = default;       // not first declaration
— end example ]

Then the question is of course which functions can be implicitly declared and when does that happen, which I have explained at:

What does "default" mean after a class' function declaration?

5 Answers