Why does "->" operator works on TSharedRef

[sleek halo]

Why does -> operator works on TSharedRef?

typedef TSharedRef<IHttpRequest, ESPMode::ThreadSafe> FHttpRequestRef

[tepid breach]

because it has the operator implemented to get the wrapped ptr?

smth like

class Foo
{
  public:
    Bar* operator->() const { return MyBar; }
  private:
    Bar* MyBar{nullptr};
};

[sleek halo]

So are you saying that shared references are not actually references?

[tepid breach]

a shared reference is a sharedptr that can't wrap a nullptr

[sleek halo]

Bruh.

[forest bone]

you can name the type as you want

and any type can have operator overloads

[sleek halo]

Thanks. This makes sense.

[narrow saddle]

is it normal that -> operator returns pointer and not object? 🤔 .
Just trying to understand if this just an example or valid case :D.
cause I think then you should do -> twice on such object?
Cause it returns member as pointer? 🤔

Foo someObject{};

someObject->()->

or am I exaggerating