Using scope operator to access non-static member variables

Question

It's a fact that you can explicitly access member variables (inside a member function, and not particularly a constructor) using this syntax : this->member_name (i.e. to distinguish with a function argument with the same name).

Besides this, I thought that the syntax ClassName::static_member was reserved to access static members outside of a class.

Then I was surprised when I realized that the following set_2() method was working as one could expect:

#include <iostream>

struct A {
    int x;
    // The two following methods seem to act similarly:
    void set_1(int x) { this->x = x; }
    void set_2(int x) { A::x = x; }
};

int main ()
{
    A a;

    a.set_1(13);
    std::cout << a.x << std::endl;

    a.set_2(17);
    std::cout << a.x << std::endl;

    return 0;
}

Output:

13
17

Is is a good and valid practice to use the scope operator (A::x) in this case? I would personnally prefer it, instead of using the this->x syntax.

IMHO, don't use names that causes conflicts like that and you don't have to use either syntax. — Some programmer dude
– Some programmer dude, Commented Jan 5, 2016 at 15:12
@JoachimPileborg : many people give this advice, but I find it quite more clear for my part to give the same name, especially with a direct assignment (only one name for one task) — yolenoyer
– yolenoyer, Commented Jan 5, 2016 at 15:17
I would decorate one of the names. I quite like trailing _ as a member variable decorator. Alternatively, you can use prefix "m_" for the member and/or "a_" for the argument. The important thing is to pick one style and stick to it. — Martin Bonner supports Monica
– Martin Bonner supports Monica, Commented Jan 5, 2016 at 15:19

Eugene · Accepted Answer · 2016-01-05 15:12:10Z

6

Using A::x in this case is valid, but I think this->x is more idiomatic and less error-prone (the reader of the code can immediately see that x is a member of the class, without thinking what A is).

answered Jan 5, 2016 at 15:12

Eugene

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Vlad from Moscow · Accepted Answer · 2016-01-05 15:15:27Z

According to the C++ Standard (3.3.7 Class scope)

2 The name of a class member shall only be used as follows:

— in the scope of its class (as described above) or a class derived (Clause 10) from its class,

— after the . operator applied to an expression of the type of its class (5.2.5) or a class derived from its class,

— after the -> operator applied to a pointer to an object of its class (5.2.5) or a class derived from its class,

— after the :: scope resolution operator (5.1) applied to the name of its class or a class derived from its class.

For example data members of methods of a derived class can hide data members and/or methods of its base class. To access data members and nethods of the base class you can use the scope resolution operator.

struct Base
{
    virtual ~Base() {};
    virtual void Hello() const { std::cout << "Base" << std::endl; }
};


struct Derived : Base
{
    virtual void Hello() const 
    { 
        Base::Hello();
        std::cout << "and Derived" << std::endl; 
    }
};

Derived d;

d.Hello();

mindriot · Accepted Answer · 2016-01-05 15:10:48Z

2

The A::x, in your case, still refers to your regular member variable; it just explicitly specifies which x you mean. Consider a class that derives from two other classes which have a member of the same name (not that that's very good coding style):

struct A { int x; };
struct B { int x; };
struct C : A, B
{
  int foo() const
  {
    // return x;  // ambiguous: which x do you mean?
    return A::x;  // unambiguous
  }
};

answered Jan 5, 2016 at 15:10

mindriot

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