When is the C ++ constructor not being called?

Question

I have a situation where the constructor is not being 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: " << ai << endl; yoyo b(5); cout << "Hello2, i: " << bi << endl; yoyo c = b; /* 1 */ cout << "Hello3, i: " << ci << endl; return 0; } 

Output:

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

(Note: there was nothing between Hello2 and Hello3)

If I changed the program 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: " << ai << endl; yoyo b(5); cout << "Hello2, i: " << bi << endl; yoyo c; c = b; /* 1 */ cout << "Hello3, i: " << ci << endl; return 0; } 

(The only difference is in the line marked with the symbol / * 1 * /)

Now the conclusion:

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

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

Answer 1

When

 yoyo c = b; 

created copy constructor .

And in case

 yoyo c; c = b; 

This is the called copy assign statement.

If you do not provide any of them, the compiler will generate default versions for you.

---

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

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

The copy assignment operator is as follows:

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

Both are 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 called copy initialization; the instance constructor created by the compiler is called, and c will be initialized with this copy. In addition, the default constructor c will be called.

 c = b; 

Here it is not an initialization, it is an appointment. On this line, the assignment operator created by the compiler will be called.

Answer 4

in your code

yoyo c=b will call the copy constructor. If you want it to be called, you must define it explicitly.

eg:

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

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

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

you can overload the assignment operator as shown below

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