4

我需要在解包和转换参数列表时获取参数的索引。有没有针对以下问题的解决方案：

#include <iostream>
#include <vector>
#include <string>

using namespace std;

void test(int a, std::string b, bool c)
{
    cout << a << "," << b << "," << c << endl ;
}

template <typename... ARG>
static void call_test(const vector<void*> &params)
{
    test(*static_cast<ARG*>(params[ indexOf(ARG) ])...);
}

int main(int argc, char **argv)
{
    int    a = 1;
    string b = "string";
    bool c   = false;
    vector<void*> v(3);
    v[0] = &a;
    v[1] = &b;
    v[2] = &c;

    call_test<int,string,bool>(v);
}
4

2 回答 2

5

我就是这样做的。首先，您将需要一些机制来创建整数的编译时序列：

using namespace std;

//===========================================================================
// META-FUNCTIONS FOR CREATING INDEX LISTS

// The structure that encapsulates index lists
template <size_t... Is>
struct index_list
{
};

// Collects internal details for generating index ranges [MIN, MAX)
namespace detail
{
    // Declare primary template for index range builder
    template <size_t MIN, size_t N, size_t... Is>
    struct range_builder;

    // Base step
    template <size_t MIN, size_t... Is>
    struct range_builder<MIN, MIN, Is...>
    {
        typedef index_list<Is...> type;
    };

    // Induction step
    template <size_t MIN, size_t N, size_t... Is>
    struct range_builder : public range_builder<MIN, N - 1, N - 1, Is...>
    {
    };
}

// Meta-function that returns a [MIN, MAX) index range
template<size_t MIN, size_t MAX>
using index_range = typename detail::range_builder<MIN, MAX>::type;

//===========================================================================

然后，您可以使用该机制来实现函数调用，利用参数包扩展的力量：

#include <iostream>
#include <vector>
#include <string>

void test(int a, std::string b, bool c)
{
    cout << a << "," << b << "," << c << endl ;
}

namespace detail
{
    // This is the function that does the real work.
    template<typename... Ts, size_t... Is>
    void call_test(const vector<void*>& params, index_list<Is...>)
    {
        test((*static_cast<Ts*>(params[Is]))...);
    }
}

// This function just creates the compile-time integer sequence and 
// forwards to another function that performs the real work.
// In other words, this is a proxy that hides the complexity of the
// machinery from the client.
template <typename... ARG>
void call_test(const vector<void*>& params)
{
    detail::call_test<ARG...>(params, index_range<0, sizeof...(ARG)>());
}

int main(int argc, char **argv)
{
    int    a = 1;
    string b = "string";
    bool c   = false;
    vector<void*> v(3);
    v[0] = &a;
    v[1] = &b;
    v[2] = &c;

    call_test<int,string,bool>(v);
}
于 2013-02-23T01:51:44.080 回答
0

最简单的解决方案肯定是不使用 a vectorof void*; 只需直接使用值：

template<typename...Arg>
void call_test(Arg const&...arg) {
  test(arg...);
}

（或该主题的一些变体。）但我想您想做的不仅仅是调用测试。

如果您真的想要索引，请使用这里所谓的“索引技巧”。你应该可以搜索到。

于 2013-02-23T01:48:50.413 回答