Definition of operator() in over.call

[barry....@gmail.com]

Overloading operator() is defined in [over.call] as:

Thus, a call x(arg1,...) is interpreted as x.operator()(arg1, ...) for a class object x of type T if T​::​operator()(T1, T2, T3) exists and if the operator is selected as the best match function by the overload resolution mechanism ([over.match.best]).

What does the second half of that even mean? What are T1, T2, T3 and what does it mean for T::operator() to exist? Why not just say that x(arg1, ...) means x.operator()(arg1, ...) and stop there?

[Bo Persson]

On 2017-06-08 17:13, barry....@gmail.com wrote:
> Overloading operator() is defined in [over.call]

> <http://eel.is/c++draft/over.call> as:

>
> Thus, a call x(arg1,...) is interpreted as x.operator()(arg1, ...)

> for a class object x of type T ifT​::​operator()(T1, T2, T3) exists

> and if the operator is selected as the best match function by the
> overload resolution mechanism ([over.match.best]

> <http://eel.is/c++draft/over.match.best>).
> <http://eel.is/c++draft/over.call#1.sentence-4>

>
>
> What does the second half of that even mean? What are T1, T2, T3 and
> what does it mean for T::operator() to exist? Why not just say that
> x(arg1, ...) means x.operator()(arg1, ...) and stop there?
>

There could be more than one operator(), so we also have to consider the
parameter types T1, T2, T3 and corresponding arguments to select the
best match.


Bo Persson

[Johannes Schaub]

The meaning of "x(...)" where x is a class-type object is defined in
http://eel.is/c++draft/over.call.object . The purpose of [over.call]
is to specify how many parameters a function call operator may have
and whether it may be a non-member function. All the rewriting and
overload resolution concerns are handled by over.match. You can call
the function constrained at [over.call] also by explicitly giving its
name, so over.call is not concerned with the rewriting rules when you
call it by using a class object as the callee (note: despite the
naming, over.call.object is not a subsection of [over.call].
over.call.object is a in a different sections-tree). I suspect that
the "Thus..." utterations in [over.oper] can therefore be made notes,
but I'm not totally sure.