Constexpr implicitly declared functions

Question

For a class of type T, the following members can be generated by the compiler, depending on the class:

• default constructor: T::T()
• copy constructor: T::T(const T&)
• move constructor: T::T(T&&)
• copy assignment operator: T& T::operator=(const T&)
• move assignment operator: T& T::operator=(T&&)

In C++14, and in C++17, what are the rules that lead to the generation of constexpr versions of these functions by the compiler?

Answer 1

The rule is simple: if the generated definition satisfies the requirements of a constexpr function, then it will be a constexpr function. For example, from C++17, [class.ctor]/7:

If that user-written default constructor would satisfy the requirements of a constexpr constructor (10.1.5), the implicitly-defined default constructor is constexpr.

The wording around implicit default constructors is describes in terms of what a "user-written default constructor" would look like. So "that user-written default constructor" means "what the compiler generates".

Similar wording exists for the copy/move constructors.

The wording is slightly more complex for the assignment operators, but it boils down to the same thing. The type must be a literal type and the assignment operators selected to do the copy/move for each subobject (non-static data member and base class) must be constexpr.