【发布时间】:2015-10-02 19:20:22
【问题描述】:
代码应该通过从字符串中提取参数来回调函数。 但是,顺序更改如下:(Visual Studio 2013 AND 2015!express)
"1 2 3 4" int, double, string, int -> 3 2 4 1
"1 2 3 4" int, double, float, int -> 4 3 2 1
编辑:它works properly in gcc 是MS Visual C++ compiler bug - 已针对 VS2013 和 VS2015 进行了测试。有谁知道解决方法? (也许使用了一些 C++14 特性?)
Edit2:我解决了它为参数添加索引并删除了元组 http://cpp.sh/9jc5
这里是示例:
void one(int i, double d, string s, int ii)
{
std::cout << "function one(" << i << ", " << d << ", " << s << ", " << ii << ");\n";
}
int main()
{
RegisterRPC<int, double, string, int>("test1", one);
DataSource* data = new DataSource("1 2 3 4");
functionarray["test1"](data);
}
以及完整的代码:
#include <stdlib.h>
#include <functional>
#include <tuple>
#include <map>
#include <iostream>
#include <istream>
#include <sstream>
#include <string>
// ------------- UTILITY---------------
template<int...> struct index_tuple{};
template<int I, typename IndexTuple, typename... Types>
struct make_indexes_impl;
template<int I, int... Indexes, typename T, typename ... Types>
struct make_indexes_impl<I, index_tuple<Indexes...>, T, Types...>
{
typedef typename make_indexes_impl<I + 1, index_tuple<Indexes..., I>, Types...>::type type;
};
template<int I, int... Indexes>
struct make_indexes_impl<I, index_tuple<Indexes...> >
{
typedef index_tuple<Indexes...> type;
};
template<typename ... Types>
struct make_indexes : make_indexes_impl<0, index_tuple<>, Types...>
{};
// ----------UNPACK TUPLE AND APPLY TO FUNCTION ---------
using namespace std;
template<class Ret, class... Args, int... Indexes >
Ret apply_helper(Ret(*pf)(Args...), index_tuple< Indexes... >, tuple<Args...>&& tup)
{
return pf(forward<Args>(get<Indexes>(tup))...);
}
template<class Ret, class ... Args>
Ret apply(Ret(*pf)(Args...), const tuple<Args...>& tup)
{
return apply_helper(pf, typename make_indexes<Args...>::type(), tuple<Args...>(tup));
}
template<class Ret, class ... Args>
Ret apply(Ret(*pf)(Args...), tuple<Args...>&& tup)
{
return apply_helper(pf, typename make_indexes<Args...>::type(), forward<tuple<Args...>>(tup));
}
// --- make tuple ---
template <typename T> T read(std::istream& is)
{
T t; is >> t; cout << t << endl; return t;
}
template <typename... Args>
std::tuple<Args...> parse(std::istream& is)
{
return std::make_tuple(read<Args>(is)...);
}
template <typename... Args>
std::tuple<Args...> parse(const std::string& str)
{
std::istringstream ips(str);
return parse<Args...>(ips);
};
// ---- RPC stuff
class DataSource
{
std::string data;
public:
DataSource(std::string s) { data = s; };
template<class...Ts> std::tuple<Ts...> get() { return parse<Ts...>(data); };
};
std::map<std::string, std::function<void(DataSource*)> > functionarray;
template<typename... Args, class F>
void RegisterRPC(std::string name, F f) {
functionarray[name] = [f](DataSource* data){apply(f, data->get<Args...>()); };
}
// --------------------- TEST ------------------
void one(int i, double d, string s, int ii)
{
std::cout << "function one(" << i << ", " << d << ", " << s << ", " << ii << ");\n";
}
int main()
{
RegisterRPC<int, double, string, int>("test1", one);
DataSource* data=new DataSource("1 2 3 4");
functionarray["test1"](data);
system("pause");
return 0;
}
// --------------------- TEST ------------------
参考文献
How do I expand a tuple into variadic template function's arguments?
【问题讨论】:
-
您希望如何使用
data->get<Args...>()区分重复的类型参数? -
@AndreyNasonov:嗯,我希望编译器能自动解决这个问题..
-
不,它做不到。调用
data->get<int>()会产生什么结果? -
@AndreyNasonov 为了安全起见,需要以某种方式使用 get
(tup) 助手? -
我认为是的,您应该使用索引而不是类型名称
标签: c++ templates tuples c++14 variadic-templates