Thread: operator overload

hi,

i got some problem understand operator overload in C++

Code:

template <class T>
class SShellPtr
{
public:
    ~SShellPtr ()
    {
        Free ();
        _malloc->Release ();
    }
    T * weak operator->() { return _p; }
    T const * operator->() const { return _p; }
    operator T const * () const { return _p; }    //what does this line mean??
    T const & GetAccess () const { return *_p; }

It declares an operator T const * for SShellPtr. No return type is required, as the return type is implied by the name of the operator. So, if x is an instance of SShellPtr<T>, then;

Code:

   T const *px = x;

or, alternatively,

Code:

    T const *px = (T const *)x;

or, even (as const applies to the thing on its left, unless it is leftmost, in which case it applies to the thing on its right)

Code:

    const T *px = (const T *)x;

In all these cases, the two lines;

Code:

    px->SomeConstMemberFunctionOfT();

and

Code:

    x->SomeConstMemberFunctionOfT();

have the same effect.