Which constructor is called first?

Question

I have looked at similar queries but I seem to be getting myself lost. I have a simple example, so please consider the following:

#include <iostream>
using namespace std;

class Animal
{
  public:
    Animal() {cout << "Animal" << endl;}
};

class Cat : public Animal
{
  public:
    Cat() {cout << "Cat" << endl;};
};

int main()
{
  Cat c;
  return 0;
}

When the program runs, it displays

Animal
Cat

My question is now this: Which constructor is actually called first. Is Cat() called and then Cat() calls Animal() before it executes its contents OR does the compiler/program look at Cat(), see that it's an Animal() and calls Animal() first then Cat()?

It's most likely that Cat is called first, and then runs the Animal constructor before printing "Cat". But if you really want to test this, set up a static variable and print then increment it to see which one runs first. — Radnyx, Aug 28 '13 at 18:42
`Cat()` is called, first thing it does it calls all base constructors in appropriate order, then `vptr` if needed (I think) and initializer list, then its own body. — lapk, Aug 28 '13 at 18:43
@PherricOxide There was a tutorial that was given to the students and I was curious cause they had two questions: which order are they called and which order are they executed? Answer to both is apparently Animal() then Cat() and I was wondering if that was the actual answer. — Metal Paw, Aug 29 '13 at 10:20

CodeClown42 · Answer 1 · 2013-08-28T18:51:15.593

4

The Animal constructor is executed before the Cat constructor body as part of the initialization of the Cat object when the Cat constructor is called. It's the same as if you had done this explicitly in an initialization list:

Cat () : Animal() {
     cout << "Cat" << endl;
}

If you want to pass arguments to a base class constructor, then you must do it explicitly as above, otherwise the default constructor (one with no parameters) is called for you. In either case, the base class constructor completes before initialization of the derived object continues.

edited Aug 28 '13 at 18:51

answered Aug 28 '13 at 18:45

CodeClown42

11,194
1
32
67

+1 good rule: `It's the same as if you had done this explicitly in an initialization list` – Wolf Jun 19 '14 at 11:14

score 1 · Answer 2 · answered Aug 28 '13 at 18:51

When the Cat constructor is called, two things happens, first the initialization list is executed and then the construction function. Implicitly you are doing this:

class Cat : public Animal
{
  public:
    Cat() 
    : Animal()
    {
        cout << "Cat" << endl;
    };
};

Then the Animal constructor is executed before the Cat implementation of the Cat constructor, but after it initialization list. For example, if the Cat and Animal classes have some members and you want to initialize them in the constructor, you could see this more easily:

class Animal
{
  private:
    bool haveHair;
  public:
    Animal(bool hair)
    :    haveHair(hair)
    {
        cout << "Animal" << endl;
    }
};

class Cat : public Animal
{
  public:
    Cat() 
    : Animal(true)
    {
        cout << "Cat" << endl;
    }
};

score 0 · Answer 3 · answered Aug 28 '13 at 18:45

0

First Animal() and then Cat() is called.

First the Animal class allocates memory and then the Cat. Maybe you want to access resources from Animal in the Cat constructor.

answered Aug 28 '13 at 18:45

DarianAnjuhal

71
2
9

`First the Animal class allocates memory` you are not talking about C++ here. – Wolf Jun 19 '14 at 11:13

Which constructor is called first?

3 Answers3