If init lists always be processed to constructor body, but it appears else in my code

Question

So, I've read this: Will the initialization list always be processed before the constructor code?

and given the following constructor:

public:
    A (int x=5):x(x+1)
    {
        cout << "In A::A x= " << x << endl;
    }

and the sample of code in the main:

A a1(10);

I don't understand the result: "In A::A x = 10" when according to my logic it should be: "In A::A x = 11"

But instead, x = 11 only after the constructor body invoked. Why is that?

Answer 1

The "problem" with this code is that both the parameter to the constructor and the member variable are named x. That is, x might not refer to the x you expect.

In this case, x refers to the parameter of the constructor - and that has a value of 10. The reason is that when C++ encounters a scope where there are two variables with the same identifier, then the most local scope wins. Here: the x from the parameter value. If you want to use the memver variable, change the code to use this.x instead of just x:

A (int x=5):x(x+1)
{
    cout << "In A::A x= " << this->x << endl;
}

Now the value you see should be 11, not 10. this is a pointer to the current object, so this->x is the value of the member variable x of the current object.

Of course, it would be still better to use different names. That way you can avoid such confusion.