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#### Detecting the Arity of Function Objects

In several places, Proto needs to know whether or not a function object `Fun` can be called with certain parameters and take a fallback action if not. This happens in `proto::callable_context<>` and in the `proto::call<>` transform. How does Proto know? It involves some tricky metaprogramming. Here's how.

Another way of framing the question is by trying to implement the following `can_be_called<>` Boolean metafunction, which checks to see if a function object `Fun` can be called with parameters of type `A` and `B`:

    template<typename Fun, typename A, typename B>
    struct can_be_called;

First, we define the following `dont_care` struct, which has an implicit conversion from anything. And not just any implicit conversion; it has a ellipsis conversion, which is the worst possible conversion for the purposes of overload resolution:

    struct dont_care
    {
        dont_care(...);
    };

We also need some private type known only to us with an overloaded comma operator (!), and some functions that detect the presence of this type and return types with different sizes, as follows:

    struct private_type
    {
        private_type const &operator,(int) const;
    };

    typedef char yes_type;      // sizeof(yes_type) == 1
    typedef char (&no_type)[2]; // sizeof(no_type)  == 2

    template<typename T>
    no_type is_private_type(T const &);

    yes_type is_private_type(private_type const &);

Next, we implement a binary function object wrapper with a very strange conversion operator, whose meaning will become clear later.

    template<typename Fun>
    struct funwrap2 : Fun
    {
        funwrap2();
        typedef private_type const &(*pointer_to_function)(dont_care, dont_care);
        operator pointer_to_function() const;
    };

With all of these bits and pieces, we can implement `can_be_called<>` as follows:

    template<typename Fun, typename A, typename B>
    struct can_be_called
    {
        static funwrap2<Fun> &fun;
        static A &a;
        static B &b;

        static bool const value = (
            sizeof(no_type) == sizeof(is_private_type( (fun(a,b), 0) ))
        );

        typedef mpl::bool_<value> type;
    };

The idea is to make it so that `fun(a,b)` will always compile by adding our own binary function overload, but doing it in such a way that we can detect whether our overload was selected or not. And we rig it so that our overload is selected if there is really no better option. What follows is a description of how `can_be_called<>` works.

We wrap `Fun` in a type that has an implicit conversion to a pointer to a binary function. An object `fun` of class type can be invoked as `fun(a, b)` if it has such a conversion operator, but since it involves a user-defined conversion operator, it is less preferred than an overloaded `operator()`, which requires no such conversion.

The function pointer can accept any two arguments by virtue of the `dont_care` type. The conversion sequence for each argument is guaranteed to be the worst possible conversion sequence: an implicit conversion through an ellipsis, and a user-defined conversion to `dont_care`. In total, it means that `funwrap2<Fun>()(a, b)` will always compile, but it will select our overload only if there really is no better option.

If there is a better option --- for example if `Fun` has an overloaded function call operator such as `void operator()(A a, B b)` --- then `fun(a, b)` will resolve to that one instead. The question now is how to detect which function got picked by overload resolution.

Notice how `fun(a, b)` appears in `can_be_called<>`: `(fun(a, b), 0)`. Why do we use the comma operator there? The reason is because we are using this expression as the argument to a function. If the return type of `fun(a, b)` is `void`, it cannot legally be used as an argument to a function. The comma operator sidesteps the issue.

This should also make plain the purpose of the overloaded comma operator in `private_type`. The return type of the pointer to function is `private_type`. If overload resolution selects our overload, then the type of `(fun(a, b), 0)` is `private_type`. Otherwise, it is `int`. That fact is used to dispatch to either overload of `is_private_type()`, which encodes its answer in the size of its return type.

That's how it works with binary functions. Now repeat the above process for functions up to some predefined function arity, and you're done.