Difference between this->field and Class::field?

Question

I'm wondering something in C++.

Admitting the following code:

int bar;
class Foo
{
public:
    Foo();
private:
    int bar;
};

Inside my class, is there any difference between this->bar and Foo::bar? Are there cases where one is invalid?

Of course, if `bar` is a virtual function instead of a data member, there's a difference between the two. Also note you can combine them like `this->Foo::bar`. — dyp, Apr 27 '13 at 22:21

score 12 · Accepted Answer · answered Apr 27 '13 at 22:13

Inside class Foo (specifically) there is no difference between the two given that bar is not static.

Foo::bar is called the fully qualified name of the member bar, and this form is useful in scenarios where there may be several types in the hierarchy defining a member with the same name. For example, you would need to write Foo::bar here:

class Foo
{
  public: Foo();
  protected: int bar;
};

class Baz : public Foo
{
  public: Baz();
  protected: int bar;

  void Test()
  {
      this->bar = 0; // Baz::bar
      Foo::bar = 0; // the only way to refer to Foo::bar
  }
};

score 4 · Answer 2 · answered Apr 27 '13 at 22:14

4

They do the same thing members.

However, you wont be able to use this-> to distinguish members of the same name in the class hierarchy. You will need to use the ClassName:: version to do that.

answered Apr 27 '13 at 22:14

pmr

58,701
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3

gcc disagrees that `this->` doesn't work with static variables: http://ideone.com/N4dSDD – Xymostech Apr 27 '13 at 22:16
@Xymostech I'm baffled. `clang` accepts it as well. Thanks for pointing it out. – pmr Apr 27 '13 at 22:18
1

No problem. I'm a bit confuzzled as well. – Xymostech Apr 27 '13 at 22:19

score 0 · Answer 3 · answered Apr 27 '13 at 22:06

0

For what I have learned fiddling around with C/C++, using the -> on something is mainly for a pointer object, and using :: is used for the classes that are part of a namespace or super class that is the general class of whatever you're including

answered Apr 27 '13 at 22:06

user2277872

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Well, for the :: it is like using std::cout << ""; Using :: means that you are using the direct implementation of the std namespace instead of saying using namespace std; in your file. The -> is like you are "pointing" a pointer to another object. That is how i have seen them used. – user2277872 Apr 27 '13 at 22:22

score 0 · Answer 4 · answered Apr 27 '13 at 22:17

It also allows you you to reference variable of another class (base class in most cases) with the same name. For me it's obvious from this example, hope it'll helps you.

#include <iostream>
using std::cout;
using std::endl;

class Base {
public:
    int bar;
    Base() : bar(1){}
};

class Derived : public Base {
public:
    int bar;

    Derived() : Base(), bar(2) {}

    void Test() {
        cout << "#1: " << bar << endl; // 2
        cout << "#2: " << this->bar << endl; // 2
        cout << "#3: " << Base::bar << endl; // 1
        cout << "#4: " << this->Base::bar << endl; // 1
        cout << "#5: " << this->Derived::bar << endl; // 2
    }
};

int main()
{
    Derived test;
    test.Test();
}

This is because the real data stored in class is like this:

struct {
    Base::bar = 1;
    Derived::bar = 2; // The same as using bar
}

Difference between this->field and Class::field?

4 Answers4