How does boost bind work behind the scenes in general?

问题

Without spending a long time reviewing the boost source code, could someone give me a quick rundown of how boost bind is implemented?

回答1:

I like this piece of the bind source:

template<class R, class F, class L> class bind_t
{
public:

    typedef bind_t this_type;

    bind_t(F f, L const & l): f_(f), l_(l) {}

#define BOOST_BIND_RETURN return
#include <boost/bind/bind_template.hpp>
#undef BOOST_BIND_RETURN

};

Tells you almost all you need to know, really.

The bind_template header expands to a list of inline operator() definitions. For example, the simplest:

result_type operator()()
{
    list0 a;
    BOOST_BIND_RETURN l_(type<result_type>(), f_, a, 0);
}

We can see the BOOST_BIND_RETURN macro expands to return at this point so the line is more like return l_(type...).

The one parameter version is here:

template<class A1> result_type operator()(A1 & a1)
{
    list1<A1 &> a(a1);
    BOOST_BIND_RETURN l_(type<result_type>(), f_, a, 0);
}

It's pretty similar.

The listN classes are wrappers for the parameter lists. There is a lot of deep magic going on here that I don't really understand too much though. They have also overloaded operator() that calls the mysterious unwrap function. Ignoring some compiler specific overloads, it doesn't do a lot:

// unwrap

template<class F> inline F & unwrap(F * f, long)
{
    return *f;
}

template<class F> inline F & unwrap(reference_wrapper<F> * f, int)
{
    return f->get();
}

template<class F> inline F & unwrap(reference_wrapper<F> const * f, int)
{
    return f->get();
}

The naming convention seems to be: F is the type of the function parameter to bind. R is the return type. L tends to be a list of parameter types. There are also a lot of complications because there are no less than nine overloads for different numbers of parameters. Best not to dwell on that too much.

回答2:

By the way, if bind_t is collapsed and simplified by including boost/bind/bind_template.hpp , it becomes easier to understand like the following :

template<class R, class F, class L> 
class bind_t
{
    public:

        typedef bind_t this_type;

        bind_t(F f, L const & l): f_(f), l_(l) {}

        typedef typename result_traits<R, F>::type result_type;
        ...
        template<class A1> 
            result_type operator()(A1 & a1)
            {
                list1<A1 &> a(a1);
                return l_(type<result_type>(), f_, a, 0);
            }
    private:
        F f_;
        L l_;

};

回答3:

I think it's a template class that declares a member variable for the arguments you want to bind and overloads () for the rest of the arguments.

来源：https://stackoverflow.com/questions/112738/how-does-boost-bind-work-behind-the-scenes-in-general

标签

c++

boost

boost-bind