How can I inspect the default copy/move constructors/assignment operators?

masher_36061 0
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How do I find out what exactly my classes' default constructors, destructors, and copy/move constructors/assignment operators do?

I know about the rule of 0/3/5, and am wondering what the compiler is doing for me.

If it matters, I'm interested in >=C++17.

c++17 c++
2 Answers

The implicitly-defined copy constructor

... performs full member-wise copy of the object's bases and non-static members, in their initialization order, using direct initialization...

For a simple structure:

struct A
{
    int x;
    std::string y;
    double z;
};

The copy-constructor would be equivalent to:

A::A(A const& otherA)
    : x(otherA.x), y(otherA.y), z(otherA.z)
{
}
user_4405580

It just calls the respective operation for every member and direct base class, nothing else.

E.g. the implicitly generated copy assignment calls copy assignment for every member.

Note that:

  • virtual bases are always initialized by the most-derived class. Any initializers for the virtual bases in the member-init-lists of any bases are ignored. Example:

    struct A
{
    int x;
    A(int x) : x(x) {}
};

struct B : virtual A
{
    B()
        : A(1) // This is ignored when constructing C.
    {}
};

struct C : B
{
    C()
        : A(2) // If A is not default-constructible, removing this causes an error.
    {}
};

C c; // .x == 2

  • The implicitly generated default constructor has a unique property (shared by a default constructor that's explicitly =defaulted in the class body): if the object was created using empty parentheses/braces, any field that would otherwise be uninitialized is zeroed. Example:

    struct A
{
    int x;
    // A() = default; // Has the same effect.
};

A f; // Uninitialized.
A g{}; // Zeroed.
A h = A{}; // Zeroed.
A i = A(); // Zeroed.

    This applies to scalar types, and the effect propagates recursively through member class instances that have the same kind of default constructors.

user_27520750
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