What is the ?-->? operator in C++?

There is no such operator in C++. Sometimes, we need to create wrapper types. For example, types like unique_ptr, shared_ptr, optional and similar. Usually, these types have an accessor member function called .get but they also provide the operator→ to support direct access to the contained value similarly to what ordinary pointers do.

The problem is that sometimes we have a few of these types nested into each other. This means that we need to call .get multiple times or to have a lot of dereference operators until we reach the value.

Something like this −

wrapper<wrapper<std::string>> wp;
wp.get().get().length();
wp.get()->length();

This can be a bit ugly. If we can replace one .get() with an arrow, it would be nice if we could replace the second .get() as well. For this, the C++98 introduced a long arrow operator.

wrapper<wrapper<std::string>> wp;
wp--->length();

What if we have another layer of wrapping? Just make a longer arrow.

wrapper<wrapper<wrapper<std::string>>> wp;
wp----->length();

The long arrow is not a single operator, but a combination of multiple operators. In this case, a normal -> operator and the postfix decrement operator --.

So, when we write wp----→length(), the compiler sees ((wp--)--)→length().

If we define the postfix -- to be the same as the dereference operator, we get the long arrow, and the even longer arrow operators −

template <typename T>
class wrapper {
   public:
   T* operator->()    { return &t; }
   T& operator--(int) {  return t; }
   private:
   T t;
};