如果不为具有不同签名的不同派生构造函数调用基类的构造函数，会发生什么情况？

Question

Say I have a base and derived class like this:

class Base {
public:
    Base() { ... };
    Base(int param) { ... };
};

class Derived : public Base {
public:
    Derived() { ... };
    Derived(int param) { ... };
    Derived(int p1, int p2) { ... };
};

Note that I'm not explicitly calling any of Base 's constructors for Derived 's constructors. That is to say, I didn't write Derived(): Base() {... } or Derived(int param): Base(param) {... } .

Do any of Base 's constructors get called by default when creating an instance of Derived ?

If so, does Base(int param) get called when using Derived(int param) , or does Base() get called?

Put another way, does C++ always default to using a base class's constructor with the same signature as the derived class's constructor if you don't specify which constructor to use? Or does it just use the base class's default constructor?

If the former, what about when using a constructor in the derived class that doesn't have a matching constructor with the same signature in the base class, such as Derived(int p1, int p2) ?

Please note this question does not relate to initialization of member variables of either class. I intentionally did not include any member variables in my pseudo-code. It specifically has to do with which constructor on the base class gets used if you do not explicitly specify a base constructor in the derived class's constructors.

Answer 1

Quoting from cppreference's description of constructors and how inheritance relates to them:

Before the compound statement that forms the function body of the constructor begins executing, initialization of all direct bases , virtual bases, and non-static data members is finished. Member initializer list is the place where non-default initialization of these objects can be specified. For bases and non-static data members that cannot be default-initialized, such as members of reference and const-qualified types, member initializers must be specified. No initialization is performed for anonymous unions or variant members that do not have a member initializer.

(Emphasis added.)

If you do not specify an initialization of the base class subobject in your derived class's constructor's member initializer list, the base class subobject is default-initialized.