Question

I saw this code but I couldn’t understand what it does:

inline S* O::operator->() const
{
    return ses; //ses is a private member of Type S*
}

so what happens now if I used ->?

Answer 1

Is you have an instance of class O and you do

obj->func()

then the operator-> returns ses and then it uses the returned pointer to call func().

Full example:

struct S
{
    void func() {}
};

class O
{
public:
    inline S* operator->() const;
private:
    S* ses;
};

inline S* O::operator->() const
{
    return ses;
}

int main()
{
    O object;
    object->func();
    return 0;
}