looking for better way to do this possibly

This is a discussion on looking for better way to do this possibly within the C++ Programming forums, part of the General Programming Boards category; In the following code: Code: #include <iostream> class Base{}; class Bar{ public: Bar() : fptr(0) {} const Base* get_origin() const ...

  1. #1
    Use this: dudeomanodude's Avatar
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    looking for better way to do this possibly

    In the following code:
    Code:
    #include <iostream>
    
    class Base{};
    
    class Bar{
        public:
            Bar() : fptr(0) {}
            const Base* get_origin() const { return fptr; }
        private:
            Base* fptr;
        friend class Foo;
    };
    
    class Foo : public Base{
        public:
            Bar make_bar(){
                Bar b;
                b.fptr = this;
                return b;
            }
    };
    
    void test_equality( const Base* f1, const Base* f2 ){
        if( f1 == f2 )
            std::cout << "Pointers are equal.\n";
        else
            std::cout << "Pointers are not equal.\n";
    }
    
    int main(){
    
        Foo f;
        Bar b  = f.make_bar();
        Bar b2 = f.make_bar();
        
        Bar c;
    
        test_equality( b.get_origin(), b2.get_origin() );
        test_equality( b.get_origin(),  c.get_origin() );
        
        return 0;
    }
    I want to make sure that a Bar instantiated on its own is never "equal" to a Bar created by a Foo by checking the Base pointer. I had to add the Base class and derive Foo from it because declaring a Foo pointer in Bar results in error( declaration of Foo not found ).

    Is there an easier/more-elegant way to achieve this kind of checking paradigm?
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  2. #2
    Cat without Hat CornedBee's Avatar
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    How about:
    Code:
    class Bar
    {
      bool createdByFoo;
      friend class Foo;
      Bar(bool) : createdByFoo(true) {}
    public:
      Bar() : createdByFoo(false) {}
    };
    
    class Foo
    {
    public:
      Bar make_bar() { return Bar(true); }
    };
    Then define the comparison appropriately.
    All the buzzt!
    CornedBee

    "There is not now, nor has there ever been, nor will there ever be, any programming language in which it is the least bit difficult to write bad code."
    - Flon's Law

  3. #3
    Use this: dudeomanodude's Avatar
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    Thanks a lot. That definitely solves one of my earlier issues, however i really want a pointer to a Foo because I don't want Bar created by different Foos to be equal either -I apologize, I realize I didn't specify that in my OP.
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  4. #4
    Cat without Hat CornedBee's Avatar
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    Then use a forward declaration instead of the ugly base class.
    Code:
    class Foo;
    class Bar
    {
      Foo *theFoo;
      friend class Foo;
      Bar(Foo *foo) : theFoo(foo) {}
    public:
      Bar() : theFoo(0) {}
    };
    
    class Foo
    {
    public:
      Bar make_bar() { return Bar(this); }
    };
    All the buzzt!
    CornedBee

    "There is not now, nor has there ever been, nor will there ever be, any programming language in which it is the least bit difficult to write bad code."
    - Flon's Law

  5. #5
    Use this: dudeomanodude's Avatar
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    Worked like a charm! I didn't go with the forward declaration because it failed in another (similar) problem I was having, so I assumed it wouldn't work for this one.

    Mucho thanks for the private constructor tip, so simple, yet I never would've thought to do it that way.
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