On Thu, May 15, 2008 at 2:07 AM, Marco Costalba <mcostalba@gmail.com> wrote:
On Thu, May 15, 2008 at 3:06 AM, Daniel Walker <daniel.j.walker@gmail.com> wrote:
I've attached a file that provides a simple unary implementation of all this as a proof of concept and to give an idea of what the boilerplate would look like. It should work for both C++03 and 0x. I tested with gcc 4.3. If there's interest, maybe this and Marco's work and possible parts of Egg could be organized into something really useful.
Hi Daniel,
thanks a lot definitely there is interest, I will take a deep look at Egg (that I don't know) and also to your polymorphic_function<>
Cool!
On this I have a naive question, just to clear the things for me.
The point of Giovanni is to use a generalized callback for type erasure of underlying function, but when you write
function<int(int,int)> f0 = functional_cast<plus(int,int)>(add_floats); function<float(float,float)> f1 = functional_cast<plus(float,float)>(add_ints);
The fact that the poly class is "plus" it's explicit anyway. IOW what's the difference of the above from
function<int(int,int)> f0 = plus(); function<float(float,float)> f1 = plus();
The assignments to f0 or f1 using functional_cast wraps a copy of the original plus object wrapped by add_floats or add_ints. The second assignments wrap brand new plus objects.
Again, in other words, does functional cast support this ?
<snip>
std::cout << f0.internal_state; // if it's 7 then you have made something very very interesting
Yes, so long as the copy constructor of the wrapped function object maintains internal_state. Alternatively, I believe you could use ref() and the wrapped object would always be the same, not a copy. Also, to access the members of the wrapped object you need to use boost::function::target() like so: std::cout << f0.target<plus>()->internal_state; functional_cast uses function::target() internally. This API is in the C++0x standard as well, so functional_cast can work with both boost::function and std::function; they're more or less the same anyway. Daniel Walker