Question

I have this sample code ( below ), example1() method works with no problem, example2() is similar but I must force const_char to make it compile, although I consider as example1() method is not needed, example2() would be not needed either.

My question is, how can I modify add() method to make both compile or how must I correctly call buffer.add() in example2() without forcing const_cast? add() method is not modifying item, so const_cast is unnecessary. Which is the correct or suitable form?

Here’s the sample code:

template <class Item>
class Buffer
{
public:
    Item *      _pItems;
    int         _nItems;
    // ... constructor / destructors etc
    void    add( const Item & item ) // or maybe Item const & item
    {
        _pItems[_nItems++] = item;
    }
};
class MyClass
{
public:
    // data
};
void    example1( const MyClass & item )
{
    Buffer<MyClass>        buffer;
    buffer.add( item );  // WORKS, no problem
}
void    example2( const MyClass & item )
{
    Buffer<MyClass *>        buffer; // NOW with pointers to MyClass
    //buffer.add( item );  // ERROR: 'Buffer<Item>::add' : cannot convert parameter 1 from 'const MyClass' to 'MyClass *const &'
    buffer.add( const_cast<MyClass *>( &item ) );  // forcing const_cast WORKS
}

Answer 1

You should do something like :

Buffer<MyClass const*> 

because &item on a const MyClass is a Myclass const* not a MyClass*