Using overloaded operators

This is a discussion on Using overloaded operators within the C++ Programming forums, part of the General Programming Boards category; Hi guys Please take a look at this example: Code: class X { public: // member prefix ++x void operator++() ...

  1. #1
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    Using overloaded operators

    Hi guys

    Please take a look at this example:

    Code:
    class X {
    public:
    
      // member prefix ++x
      void operator++() { }
    };
    
    class Y { };
    
    // non-member prefix ++y
    void operator++(Y&) { }
    
    int main() {
      X x;
      Y y;
    
      // calls x.operator++()
      ++x;
    
      // explicit call, like ++x
      x.operator++();
    
      // calls operator++(y)
      ++y;
    
      // explicit call, like ++y
      operator++(y);
    }
    Please note that when using the overloaded operator for an instance of X, we can call it using

    Code:
    x.operator++()
    In the case with the overloaded operator ++ for an instance of Y (it is overloaded using a non-member function), we can call it using

    Code:
    operator++(y)
    My question is: Is there any reasoning behind the fact that we cannot call the two operators using the same syntax? Or is that just the way it is?

    Best,
    Niles.

  2. #2
    C++ Witch laserlight's Avatar
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    Quote Originally Posted by Niels_M
    Is there any reasoning behind the fact that we cannot call the two operators using the same syntax?
    But you can, i.e., ++x and ++y

    Besides that, please take a look at this example:
    Code:
    class X {
    public:
    
      // member function foo
      void foo() { }
    };
    
    class Y { };
    
    // non-member function foo
    void foo(Y&) { }
    
    int main() {
      X x;
      Y y;
    
      x.foo();
    
      foo(y);
    }
    Please note that when using the function foo for an instance of X, we can call it using
    Code:
    x.foo();
    In the case with the function foo for an instance of Y (it is declared as a non-member function), we can call it using
    Code:
    foo(y);
    My question is: Is there any reasoning behind the fact that we cannot call the functions using the same syntax? Or is that just the way it is?
    Quote Originally Posted by Bjarne Stroustrup (2000-10-14)
    I get maybe two dozen requests for help with some sort of programming or design problem every day. Most have more sense than to send me hundreds of lines of code. If they do, I ask them to find the smallest example that exhibits the problem and send me that. Mostly, they then find the error themselves. "Finding the smallest program that demonstrates the error" is a powerful debugging tool.
    Look up a C++ Reference and learn How To Ask Questions The Smart Way

  3. #3
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    Quote Originally Posted by laserlight View Post
    But you can, i.e., ++x and ++y
    Ah, yes. Of course.


    Quote Originally Posted by laserlight View Post
    Besides that, please take a look at this example:
    Code:
    class X {
    public:
    
      // member function foo
      void foo() { }
    };
    
    class Y { };
    
    // non-member function foo
    void foo(Y&) { }
    
    int main() {
      X x;
      Y y;
    
      x.foo();
    
      foo(y);
    }
    Please note that when using the function foo for an instance of X, we can call it using
    Code:
    x.foo();
    In the case with the function foo for an instance of Y (it is declared as a non-member function), we can call it using
    Code:
    foo(y);
    I see, very good explanation. Thanks!

    Best,
    Niles.

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