From: Steven Watanabe (watanabesj_at_[hidden])
Date: 2008-03-22 14:36:54
AMDG
Markus Werle wrote:
> Now we introduce another function call - not an operator
> this time, but in C++ these are equivalent.
>
> struct fun_t {};
> terminal<fun_t>::type const fun = {{}};
>
> fun(a, b);
>
> This is represented by something similar to
>
> expr<
> tag::function
> ,args3<
> ref_<expr<tag::terminal, args0<fun_tag> > const>
> ,expr<tag::terminal, args0<> >
> ,expr<tag::terminal, args0<> >
> >
> >
>
> This is suprising. In contrast to operator+, the function fun
> is treated as if it was an *operand*, not an *operator*.
> The expression is not tagged by _the_ function call, but by a
> global this-is-a-function-tag and the function is stored in the
> argument typelist.
>
I don't find it surprising. This behavior is exactly what I would expect.
fun(a, b) is equivalent to fun.operator()(a, b). In a concept fun(int, int)
will be represented as
concept Callable<class T> {
void operator()(T, int, int);
};
if I recall correctly.
> What fun means is context dependent, I would expect the context
> to take care about what fun_t might mean, so if we want
> symmetry first I'd expect a type representation similar to
>
> expr<
> tag::function<fun_t>
> ,args2<
> ,expr<tag::terminal, args0<...ommitted...> >
> ,expr<tag::terminal, args0<...ommitted...> >
> >
> >
>
>
> So I wonder whether this asymmetry was introduced due to the
> fact that function objects could have a state which makes them
> an in-between between operator and operand ...
>
IMO, a function object is not an operator at all. The operator
involved is the function call operator which takes a reference to
the function object (*this) as the first parameter. I don't see this
as being any more asymmetric than any other member function call.
In Christ,
Steven Watanabe