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;
}

What's the meaning of the & ?

int & getNumber()const ;

Just remove it in the function and prototype.

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.

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.

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.

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.

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:
        Base( int number = 0 )
        {
            m_iNumber = number;
        }

        void setNumber( int number )
        {
            m_iNumber = number;
        }

        const int& getNumber() const
        {
            return m_iNumber;
        }

        int& getNumber()
        {
            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. 


---

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.

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.