Question

Consider this code:

class First
{
public:
    int y;
    First()
    {
        y = 90;
    }
};

class Second{
    public:
       int x;
       First ob;  //if i comment it, value of x is initialised to zero
};

int main()
{
    Second obj = Second();
    cout << obj.x << endl;
}

This program gives different output when I change it:

1. If I comment out the line First ob; then value of x is initialised to zero
2. If the class consists object of class First, then member has garbage value.
3. If I create object like Second obj; then it has garbage value.

What is the reason it behaves differently when Second class consists only built-in members and when object of another class?

And what is the difference between these statements:

Second obj = Second();
Second obj;

Answer 1

Case 1. When you comment out the line First ob, then Second class becomes a POD type, which can be value-initialized using the syntax Second obj= Second(), and value-initialization, in case of built-in types, means zero-initialized, so x is zero-initialized.

Case 2: When you keep the line First ob, then Second class becomes non-POD type, because First class is non-POD type (as it has user-defined constructor¹]). In this case, the syntax Second obj=Second() doesn’t initialize built-in data type, if you don’t initialize it in the constructor.

Case 3: When write Second obj, then there are again two cases (read them carefully):

• obj will be default constructed if there is First ob line. And in this case, x will not be initialized at all, as it is not manually initialized in the constructor of Second.

• obj will not be initialized at all if you First ob line is commented out. In this case, Second becomes a POD, and POD types are not initialized if you only write Second obj.

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¹ See this related topic:

• Can't C++ POD type have any constructor?