I was under impression that it's impossible, see for example:
Calling the constructor of the base class after some other instructions in C++
But the following program runs and produces two lines of "Constructor Person":
#include <iostream>
class Person
{
public:
Person()
{
std::cout << "Constructor Person" << std::endl; }
};
class Child : public Person
{
public:
Child()
{
c = 1;
Person();
}
int c;
};
int main()
{
Child child;
return 0;
}
The first one is implicit call of the default constructor, that's clear. What about the 2nd one - does it mean that the action described in the title is legitimate? I use Visual C++ 2010.
The call inside the child class constructor is not calling the base class constructor, it is creating a temporary, unnamed and new object of type Person. It will be destroyed as the constructor exits. To clarify, your example is the same as doing this:
Child() { c = 1; Person tempPerson; }
Except in this case, the temporary object has a name.
You can see what I mean if you modify your example a little:
class Person
{
public:
Person(int id):id(id) { std::cout << "Constructor Person " << id << std::endl; }
~Person(){ std::cout << "Destroying Person " << id << std::endl; }
int id;
};
class Child : public Person
{
public:
Child():Person(1) { c = 1; Person(2); }
int c;
};
int main() {
Child child;
Person(3);
return 0;
}
This produces the output:
Constructor Person 1
Constructor Person 2
Destroying Person 2
Constructor Person 3
Destroying Person 3
Destroying Person 1