c++ - 通过转发构造函数参数构建基于可变参数模板的mixin

Question

我正在尝试构建一个 mixin 模板，其基数全部作为可变模板参数传递。我想通过将每个 mixin 类的构造函数参数作为参数传递给可变模板构造函数来构造 mixin

可变参数模板构造函数在调用每个混合类类型的对象时编译。但是如果我传递每个 mixin 类的构造函数参数(所有类都有一个参数构造函数)，它不会编译

我正在使用带有 -std=c++1z 的 gcc 7.0

我做错了什么？

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

 template < typename... T >
 struct Mixin : T...
 {
     Mixin() = delete;
     Mixin(Mixin const &) = delete;
     Mixin(Mixin &&) = delete;

     template < typename... U >
     Mixin(U &&... v) : T(std::forward < U >(v))...
     {
     }
 };

 int main()
 {
     using A = std::vector < std::string >;
     using B = std::unordered_map < std::string, std::string >;
     using C = std::string;
     using M = Mixin < A, B, C >;

     //  This doesn't compile
     M m1{{"hello", "world"}, { {"hello",  "world" }, {"world",  "hello"} }, "hello"};

     //  This compiles
     A a({"hello", "world"}); B b({ {"hello",  "world" }, {"world",  "hello"} }); C c("hello");
     M m2{a, b, c};
 }

Answer 1

这里的问题是 std::initializer_list 不能从 forwarding-reference 推导出来。事实上，显式指定 std::initializer_list 可以编译您的代码:

M m1{
    std::initializer_list<std::string>{"hello", "world"}, 
    std::initializer_list<std::pair<const std::string, std::string>>{{"hello",  "world" },{"world",  "hello"} }, 
    "hello"};

wandbox example

您可以找到有关std::initializer_list 和推导的更多信息in this question .

---

您可以通过创建辅助 make_il 函数强制推导 std::initializer_list:

template <typename... Ts>
auto make_il(Ts&&... xs) 
{ 
    return std::initializer_list<std::common_type_t<Ts...>>{
        std::forward<Ts>(xs)...}; 
}

您的最终代码将如下所示:

using namespace std::literals::string_literals;          
using kvp = std::pair<const std::string, std::string>;

M m1{
    make_il("hello"s, "world"s), 
    make_il(kvp("hello"s,  "world"s), kvp("world"s,  "hello"s)), 
    "hello"};

wandbox example