type & getReference() const doesn't work.

This is a discussion on type & getReference() const doesn't work. within the C++ Programming forums, part of the General Programming Boards category; The following code doesn't compile with gcc 2.96 for linux. The error I receive is : ../src/text.cxx: In method 'int ...

  1. #1
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    type & getReference() const doesn't work.

    The following code doesn't compile with gcc 2.96 for linux.
    The error I receive is :

    ../src/text.cxx: In method 'int & Base::getNumber() const':
    ../src/text.cxx:line7: could not convert 'this->Base::number' to 'int&'

    The problem vanish if the getNumber method is declared to return const:

    const int & getNumber() const;


    Is there a bug in gcc or I miss something?

    Best Regards,
    Cristian


    include <iostream>

    class Base {
    public:
    Base();
    public:
    int number;
    void setNumber(int number);
    int & getNumber()const ;
    };
    using std::cout;
    Base::Base() {
    cout << "\n\n\n\ncalling the ctrctor." << endl;
    }
    int & Base::getNumber()const {
    return number;
    }
    void Base::setNumber(int number) {
    this->number=number;
    }

    int main() {
    Base * pBase = new Base();
    pBase->setNumber(8);
    cout << "getNumber() called: " << pBase->getNumber() << endl;
    return 0;
    }

  2. #2
    Me want cookie! Monster's Avatar
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    What's the meaning of the & ?

    int & getNumber()const ;

    Just remove it in the function and prototype.

  3. #3
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    Hi,

    The function tries to return a reference to the "number" variable.

    The code I have posted is just an example of the problem I have. The real code should take a reference to the variable.

    To make me clear: I want to have a const function that returns a reference and it doesn't work. Legally it must, in my opinion.


    BR,
    Cristian

    PS;
    also const int & getNumber() const work. But I wold like to know why the first option does't work.

  4. #4
    ¡Amo fútbol!
    Join Date
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    int & Base::getNumber()const {
    return number;
    }

    should be

    int & Base::getNumber()const {
    return &number;
    }

    BTW, next time use code tags. It makes your code easier to read.

  5. #5
    endo_at_work
    Guest
    if you want to return a reference from a const function, it has to be a const reference. If not, then you are effectively able to modify a data member through a const member function - which is what they are used to prevent.

  6. #6
    Skunkmeister Stoned_Coder's Avatar
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    Your compiler is complaining probably quite rightly so....
    Why when you declare a function const (as in wont alter objects state) do you think that the compiler should allow you to return a non const reference. This would invalidate the const specification of the function.
    Free the weed!! Class B to class C is not good enough!!
    And the FAQ is here :- http://faq.cprogramming.com/cgi-bin/smartfaq.cgi

  7. #7
    the hat of redundancy hat nvoigt's Avatar
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    Hannover, Germany
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    You cannot give a reference to something unmodifiable ( const ) to someone who might modify it.

    PHP Code:
    #include <iostream>

    using namespace std;

    class 
    Base 
    {
        public:
            
    Baseint number )
            {
                
    m_iNumber number;
            }

            
    void setNumberint number )
            {
                
    m_iNumber number;
            }

            const 
    intgetNumber() const
            {
                return 
    m_iNumber;
            }

            
    intgetNumber()
            {
                return 
    m_iNumber;
            }

        private:
            
    int m_iNumber;

    };

    int main() 
    {
        
    Base base(8);

        
    cout << base.getNumber() << endl// const version called

        
    base.getNumber() = 666// non-const version called, will not compile on const objects
        
    cout << base.getNumber() << endl// const version called

        
    cin >> base.getNumber(); // non-const obviously
        
    cout << base.getNumber() << endl// const version called

        
    return 0;
    }

    // As you have both a const and non const accessor function,
    // I would suggest dropping the set function and renaming both
    // getNumber functions to Number instead.
    //
    // In C++.NET this is called a property. 
    hth
    -nv

    She was so Blonde, she spent 20 minutes looking at the orange juice can because it said "Concentrate."

    When in doubt, read the FAQ.
    Then ask a smart question.

  8. #8
    Registered User
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    Hi,

    My view was that const just speaks about what the function does and const type indicates how the result behaves. Now it's ok.

    Thank you.

  9. #9
    the hat of redundancy hat nvoigt's Avatar
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    Hannover, Germany
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    3,139
    Indeed. However, the behaviour of a const function includes throwing compiler errors if a user tries to circumwent it's meaning

    By using a const function, you should not be able to alter the state of the object. If this const function returned a direct reference to internal data, it would leave the object open to modification.
    hth
    -nv

    She was so Blonde, she spent 20 minutes looking at the orange juice can because it said "Concentrate."

    When in doubt, read the FAQ.
    Then ask a smart question.

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