const_iterator和模板，编译错误

Question

I can't understand why the following code is not working, any idea ?

template <class T>
class Matrice
{
public:
...
    typedef typename std::vector<T>::const_iterator const_iterator;
    const_iterator& cend ( )
    {
        return valeurs.cend ( );
    }
...
private:
...
}

here's the complilator's complaint :

/Users/Aleks/Documents/DS OO/DS OO/Matrice.h:70:16: Non-const lvalue reference to type 'const_iterator' (aka '__wrap_iter') cannot bind to a temporary of type 'const_iterator' (aka '__wrap_iter')

Answer 1

valeurs.cend ( cppreference ) returns an instance to a const_iterator (that is, it's declared as const_iterator valeurs.cend() ).

The compiler needs to create a temporary object (memory area) to store the value returned by valeurs.cend() . This code fails to compile because you cannot take the reference of a temporary as the latter won't outlive the function call.

You'd usually return an iterator by value:

typedef typename std::vector<T>::const_iterator const_iterator;
const_iterator cend ( )
{
    return valeurs.cend ( );
}

This will make sure the value returned by valeurs.cend() is copied (or moved, in C++11, I believe) to its the destination object (if you're assigning the returned value to a variable of type const_iterator ) or in another temporary wherever Matrice<T>::cend() is called. See the link to MSDN's explanation for details.

Answer 2

hmjd is right, you need just const_iterator, not a reference. The reason you can't use a reference is that valeurs.cend () is a temporary value on the stack, the reference (if you could use it) wouldn't be valid as soon as the function returns.

Answer 3

As others have said, the following line:

const_iterator& cend ( )

Needs to be either:

const const_iterator& cend ( )

Or:

const_iterator cend ( )