我假设您感兴趣的类型是int 和Foo,如果这不是您所追求的,请忽略此答案。查看文档,我无法找到获取这些类型的简单方法。但是,如果您查看存储在 f 中的 proto 表达式的类型,您会发现 int 和 Foo 可以在第一个孩子内的演员向量中找到。您可以在输出中看到最终获得有趣类型所需的步骤,然后您可以轻松创建一个执行您想要的元函数。在这个简单的例子中,get_local_type 使用索引来访问有问题的类型。如果您想通过名称访问它(使用_a),您应该能够使用lambda 表达式的第二个子项中map_local_index_to_tuple 中的数据获取与名称关联的索引。使用定义为here 的phoenix::detail::get_index 实现get_local_type_from_name 也很容易。这个元函数需要上面提到的地图作为它的第一个参数,以及你想要从中获取信息的占位符的类型(更具体地说,它需要phoenix::detail::local<phoenix::local_names::_a_key>,你可以在占位符的类型上使用proto::result_of::value)作为它的第二个参数。
#include <iostream>
#include <typeinfo>
#include <string>
#include <cxxabi.h>
#include <type_traits>
#include <boost/proto/proto.hpp>
#include <boost/phoenix.hpp>
namespace proto = boost::proto;
namespace phoenix = boost::phoenix;
using namespace phoenix::local_names;
namespace fusion = boost::fusion;
struct Foo { const char str[6] = " Ok.\n"; };
std::string demangle(const char* mangledName) {
int status;
char* result = abi::__cxa_demangle(mangledName, nullptr, nullptr, &status);
switch(status) {
case -1:
std::cerr << "Out of memory!" << std::endl;
exit(1);
case -2:
return mangledName;
case -3: // Should never happen, but just in case?
return mangledName;
}
std::string name = result;
free(result);
return name;
}
template <typename Lambda, int N>
struct get_local_type
{
typedef typename proto::result_of::value<typename proto::result_of::child_c<Lambda,0>::type >::type vector_of_locals_type;
typedef typename proto::result_of::value<typename fusion::result_of::at_c<vector_of_locals_type,N>::type >::type ref_type;
typedef typename std::remove_reference<ref_type>::type type;
};
template <typename Lambda, typename Arg>
struct get_local_type_from_name
{
typedef typename proto::result_of::value<Arg>::type local_name;
typedef typename proto::result_of::value<typename proto::result_of::child_c<Lambda,1>::type >::type map_type;
typedef typename phoenix::detail::get_index<map_type,local_name>::type index;
typedef typename get_local_type<Lambda,index::value>::type type;
};
int main(int argc, char *argv[])
{
auto f = phoenix::lambda(_b = 17, _a = Foo()) [
std::cout << _b << phoenix::bind(&Foo::str,_a)
];
std::cout << std::endl << "This is the whole lambda expression:" << std::endl;
std::cout << std::endl << demangle(typeid(f).name()) << std::endl;
std::cout << std::endl << "Take the first child:" << std::endl;
std::cout << std::endl << demangle(typeid(proto::child_c<0>(f)).name()) << std::endl;
std::cout << std::endl << "Then its value (this is a vector that contains the types you want):" << std::endl;
std::cout << std::endl << demangle(typeid(proto::value(proto::child_c<0>(f))).name()) << std::endl;
std::cout << std::endl << "Take the first element of that vector:" << std::endl;
std::cout << std::endl << demangle(typeid(fusion::at_c<0>(proto::value(proto::child_c<0>(f)))).name()) << std::endl;
std::cout << std::endl << "Take the value of that element:" << std::endl;
std::cout << std::endl << demangle(typeid(proto::value(fusion::at_c<0>(proto::value(proto::child_c<0>(f))))).name()) << std::endl;
typedef typename proto::tag_of<decltype( f )>::type tag;
typedef typename proto::tag_of<decltype(proto::child_c<2>(f))>::type tagc;
static_assert(proto::arity_of<decltype(f)>::value==3,"");
static_assert(std::is_same<tag, phoenix::tag::lambda_actor>::value,"");
static_assert(std::is_same<tagc, proto::tag::shift_left>::value,"");
typedef typename get_local_type<decltype(f),0>::type type_of_1st;
typedef typename get_local_type<decltype(f),1>::type type_of_2nd;
typedef typename get_local_type_from_name<decltype(f),_a_type>::type type_of_a;
typedef typename get_local_type_from_name<decltype(f),decltype(_b)>::type type_of_b;
static_assert(std::is_same<type_of_1st,int>::value,"");
static_assert(std::is_same<type_of_2nd,Foo>::value,"");
static_assert(std::is_same<type_of_a,Foo>::value,"");
static_assert(std::is_same<type_of_b,int>::value,"");
return 0;
}