When is a c++ constructor not called?

Question

I have a situation where no constructor appears to be called:

#include <iostream>

using namespace std;

int main ()
{
    class yoyo
    {
        public:
        int i;
        yoyo()
        {
            i = 0;
            cout << "defaultly initialized to 0" << endl;
        }
        yoyo (int j) : i(j)
        {
            cout << "initialized to " << j << endl;
        }
    };

    int i;

    yoyo a;
    cout << "Hello1, i: " << a.i << endl;

    yoyo b(5);
    cout << "Hello2, i: " << b.i << endl;

    yoyo c = b;                                   /* 1 */
    cout << "Hello3, i: " << c.i << endl;

    return 0;
}

Output is:

defaultly initialized to 0
Hello1, i: 0
initialized to 5
Hello2, i: 5
Hello3, i: 5

(Note: nothing between Hello2 and Hello3)

If I change the program to read as follows:

#include <iostream>

using namespace std;

int main ()
{
    class yoyo
    {
        public:
        int i;
        yoyo()
        {
            i = 0;
            cout << "defaultly initialized to 0" << endl;
        }
        yoyo (int j) : i(j)
        {
            cout << "initialized to " << j << endl;
        }
    };

    int i;

    yoyo a;
    cout << "Hello1, i: " << a.i << endl;

    yoyo b(5);
    cout << "Hello2, i: " << b.i << endl;

    yoyo c; c = b;                                  /* 1 */
    cout << "Hello3, i: " << c.i << endl;

    return 0;
}

(The only difference is in he line marked by /* 1 */)

The output now is:

defaultly initialized to 0
Hello1, i: 0
initialized to 5
Hello2, i: 5
defaultly initialized to 0
Hello3, i: 5

Now there is a constructor call between Hello2 and Hello3. My question is, why is there no (visible) constructor call in the first case?

Answer 1

In the case of

yoyo c = b;

it's the copy constructor that is called.

And in the case of

yoyo c; c = b;

it's the copy assignment operator that is called.

If you don't provide any of them, the compiler will generate default versions for you.

---

If you want to create your own copy constructor, it could look like this:

yoyo(const yoyo& other)
    : i(other.i)
    { std::cout << "copy constructor initialized\n"; }

The copy assignment operator looks like this:

yoyo& operator=(const yoyo& other)
    {
        i = other.i;
        return *this;
    }

Both of them defined inside the class definition of course.

Answer 2

In the first case:

yoyo c = b;   

calls the copy constructor , which the compiler generates for you implicitly in this case.

Answer 3

 yoyo c = b; 

This is referred to as copy-initialization ; the compiler-generated copy constructor will be called and c will be initialized with that copy. Also, the default constructor of c will be called.

c = b;

Here, this is not initialization, this is assignment. The compiler-generated assignment operator will be invoked on that line.

Answer 4

in your code,

yoyo c=b will call copy constructor.if you want to see it being called,you have to define it explicitely.

eg:

yoyo(const yoyo& obj)
{
   this->i=obj.i;  cout<<"copy constructor"<<endl;
}

in the second case it will call constructor and then the assignment operator

yoyo c; //constructor
c = b;  //assignment operator for which only copying occurs

you can overload assignment operator as below

yoyo& operator=(yoyo& obj)
{
    i = obj.i;
    cout << "assignment operator" << endl;
}