提振.变体,提振.MPL:如何追加类型

Question

我看一下基于boost的栅格代码。我不禁想知道我们是否可以使用Boost。变体。我想知道这样的API是否可能:

void voidFunc()
{
    std::cout << "void called" << std::endl;
}
int stringFunc(std::string str)
{
    std::cout << str << std::endl;
    return 0;
}
int main()
{
    some_map_like_type<std::string, boost::variant> funcs;
    funcs.insert<void , void >("voidFunc", &voidFunc)); // now our variant vould contain something like boost::function<void, void>
    funcs.insert<int , std::string>("stringFunc", &stringFunc)); // and now we added to our variant a new type: boost::function<int , std::string>
    funcs.insert<void , void >("voidFunc2", &voidFunc)); // and now our variant should not change because it already contains boost::function<void, void> type

    // And here when all the fun part is:
    funcs["voidFunc"](); // compiles
    funcs["stringFunc"]("hello"); // compiles
    funcs["stringFunc"](some_not_std_string_class); // does not compile.
    return 0;
}

这意味着最后编译器必须编译如下内容:

void voidFunc()
{
    std::cout << "void called" << std::endl;
}
int stringFunc(std::string str)
{
    std::cout << str << std::endl;
    return 0;
}
int main()
{
    some_map_like_type<std::string, boost::variant< boost::function<void , void>, boost::function<int , std::string> > > funcs;
    funcs.insert<void , void >("voidFunc", &voidFunc)); // now our variant vould contain something like boost::function<void, void>
    funcs.insert<int , std::string>("stringFunc", &stringFunc)); // and now we added to our variant a new type: boost::function<int , std::string>
    funcs.insert<void , void >("voidFunc2", &voidFunc)); // and now our variant should not change because it already contains boost::function<void, void> type

    // And here when all the fun part is:
    funcs["voidFunc"](); // compiles
    funcs["stringFunc"]("hello"); // compiles
    funcs["stringFunc"](some_not_std_string_class); // here it would give error and would not compile
    return 0;
}

更新:

我尝试了什么(基于这个Variant文档和这个MPL演示和文档):

#include <boost/static_assert.hpp>
#include <boost/mpl/equal.hpp>
#include <boost/mpl/vector_c.hpp>
#include <boost/mpl/transform.hpp>
#include <boost/mpl/multiplies.hpp>
#include <boost/mpl/placeholders.hpp>
#include <boost/variant.hpp>
#include <iostream>
#include <string>
#include <vector>
class sudo_science
{
public:
    typedef  boost::mpl::vector_c<int> types_vector1;
    typedef boost::make_recursive_variant< types_vector1 >::type recursive_variant_t;
    std::vector< recursive_variant_t > variant_seq;
    template <typename T>
    void append(T val)
    {
        typedef  boost::mpl::push_back<types_vector1,T>::type types_vector1;
        variant_seq.push_back(val);
        return;
    }
    std::vector< recursive_variant_t > give_me_end_variant()
     {
         return variant_seq;
     }
};
int main()
{
    sudo_science a;
    a.append<float>(1.0);
    a.append<std::string>("Stack and Boost");
    //sorry for C++11
    auto varint = a.give_me_end_variant();
    return 0;
}

但是编译失败，有两个相同的错误:

Error   1   error C2665: 'boost::detail::variant::make_initializer_node::apply<BaseIndexPair,Iterator>::initializer_node::initialize' : none of the 2 overloads could convert all the argument types    c:program filesboostincludeboostvariantvariant.hpp    1330    1

Answer 1

不可能。operator[]是运行时的东西，而类型是编译时的东西。那么编译器应该编译以下代码吗?

char const* str;
if (some_condition())
  str = "voidFunc";
else
  str = "stringFunc";
// ... some more code
if (some_condition())
  funcs[str]();
else
  funcs[str](str);

编译器应该如何知道第二次调用some_condition()是否给出与以前相同的结果?还是中间代码修改了str的值?

void call(some_map_like_type<std::string, boost::variant> const& funcs)
{
  funcs["voidFunc"]();
}

编译器应该如何知道是否在调用时funcs包含一个条目映射"voidFunc"到一个没有参数的函数?如果它被调用一次，它的值是，而另一次，它的值不是，会发生什么?

根据您实际想要实现的内容，可能有一种方法可以使用模板和constexpr函数来获得它。但是请注意，运行时发生的任何事情都不会影响代码是否编译，原因很简单，代码在编译之前不能运行。