Is there a way to deduce the signature of a lambda as an mpl sequence?

问题

Is there a way to deduce the signature, result- and parameter-types, of a c++0x lambda as a Boost.MPL sequence, for example a boost::mpl::vector? For example, for a lambda

[]( float a, int b ) -> void { std::cout << a << b << std::endl; }

I would like to get a boost::mpl::vector<void,float,int>.

回答1:

C++0x lambdas which are "closure-objects" are functors. So you can use boost.Boost.FunctionTypes to decompose its operator().

Example:

#include <boost/function_types/parameter_types.hpp>

#include <boost/mpl/at.hpp>
#include <boost/mpl/int.hpp>

int main()
{
    int x = 1;
    auto f = [x](char a, short b, int c){ return x; };

    typedef decltype(f) lambda_t;
    typedef boost::function_types::parameter_types<
        decltype(&lambda_t::operator())>::type args_t;
    // we can use boost::mpl::identity<decltype(f)>::type instead of lambda_t

    static_assert(sizeof(boost::mpl::at<args_t, boost::mpl::int_<1>>::type) == 1, "");
}

来源：https://stackoverflow.com/questions/4482646/is-there-a-way-to-deduce-the-signature-of-a-lambda-as-an-mpl-sequence