Question

If I have a Foo *foo, I can say foo->bar(). Is it possible to call the operator->() function manually? And if so, how would I pass it bar()?

Does it make a difference if it is Foo foo instead?

Maybe something like foo.operator->(bar)?

Answer 1

Yes, you can. With overloaded -> the foo->bar() expression is interpreted by the compiler as foo.operator->()->bar(). And this is exactly how you can call it “manually”: foo.operator->()->bar().

If your overloaded operator -> function is implemented “properly”, i.e. it returns something that also supports operator -> then there’s not much point in using the “manual” syntax, since it is doing the same thing as the “non-manual” one.

The only case you’d need the “manual” syntax is when your implementation of overloaded operator -> returns something that does not support another application of ->. An int value, for example.