Question

Suppose I have:

Foo foo;

is there a shorthand for this?

foo.operator->().operator()(1, 2);

Answer 1

Assuming you actually meant foo.operator->().operator()(1, 2), and that you have control over the class Foo, a simpler form would be (*foo)(1, 2). It requires the operator* to that defined though, but since we usually expect foo->bar to be equivalent to (*foo).bar, it seems reasonable.

If your Foo is a smart pointer class of some sort, which points to an object which defines an operator(), this would be the most concise way of calling the object’s operator().

But without more detail (and without you providing an expression that’s actually valid C++ — there’s no way in which operator(1, 2) as you wrote it can be valid), it’s impossible to answer your question. I’m just guessing at what you’re trying to do.