问题描述
我试图理解这段代码 并且我在此主题上找到了更多 SO 内容.
I am trying to understand this code and I have found some more SO content on this topic.
紧凑形式:
#include <tuple>
namespace sqlite {
namespace utility {
template<typename> struct function_traits;
template <typename Function>
struct function_traits : public function_traits<
decltype(&Function::operator())
> { };
template <
typename ClassType,
typename ReturnType,
typename... Arguments
>
struct function_traits<
ReturnType(ClassType::*)(Arguments...) const
> {
typedef ReturnType result_type;
template <std::size_t Index>
using argument = typename std::tuple_element<
Index,
std::tuple<Arguments...>
>::type;
static const std::size_t arity = sizeof...(Arguments);
};
}
}
这让我了解了该语言的指向成员函数的指针特性，并且在这一点上我很清楚如何使用函数特征(很简单，我不仅可以提取返回类型，还可以提取参数甚至参数的类型)，但我一直在理解 为什么 这有效，或者为什么它甚至可以工作..
This led me to learn about the pointer-to-member-function feature of the language, and it's pretty clear to me at this point how to use the function traits (it's fairly straightforward that I can pull out not only the return type but the arity and even the types of the arguments) but I have gotten stuck in understanding why this works, or indeed why it is even possible for it to work...
推荐答案
让我们一点一点地分解一切.
Let's break everything down piece by piece.
template<typename> struct function_traits;
这是主要的模板声明，使用单个模板参数声明模板类.
That is the primary template declaration, declaring a template class with a single template parameter.
template <typename Function>
struct function_traits : public function_traits<
decltype(&Function::operator())
> { };
这本质上是一个转发特性，允许函子(具有 operator() 的对象)与这个特性类一起使用.它是从 operator() 的特征继承的主模板的定义.当后面定义的专业不匹配时会选择这个版本.
That is essentially a forwarding trait to allow functors (objects which have an operator()) to be used with this traits class. It's a definition of the primary template which inherits from the traits for operator(). This version will be selected when the specialization which is defined later is not matched.
template <
typename ClassType,
typename ReturnType,
typename... Arguments
>
struct function_traits<
ReturnType(ClassType::*)(Arguments...) const
>
这开始定义 function_traits 的部分特化，用于指向成员函数的指针.当您将成员函数指针类型作为模板参数传递给 function_traits 时，将选择此特化.返回类型、类类型和参数类型将被推导出为模板参数.
That begins the definition of a partial specialization of function_traits for pointers to member functions. When you pass a member function pointer type as the template argument to function_traits, this specialization will be selected. The return type, class type and argument types will be deduced as template parameters.
typedef ReturnType result_type;
这是一个简单的typedef，它将推导出的ReturnType 模板参数别名为result_type.
That is a simple typedef which aliases the deduced ReturnType template parameter to result_type.
template <std::size_t Index>
using argument = typename std::tuple_element<
Index,
std::tuple<Arguments...>
>::type;
这就是所谓的别名模板.当您提供诸如 function_traits<myFunc>::template argument<2> 之类的索引时，它将折叠为 typename std::tuple_element<2, std::tuple<Arguments...>>::type.这样做的目的是在索引 Index 处提取参数的类型.作者没有编写所有模板元编程代码来手动执行此操作，而是选择使用 std::tuple 中的现有代码.std::tuple_element 提取元组的 n 个元素的类型.
That is what is called an alias template. When you provide an index such as function_traits<myFunc>::template argument<2>, it will collapse down to typename std::tuple_element<2, std::tuple<Arguments...>>::type. The goal of that is to extract the type of the argument at index Index. Rather than writing out all the template metaprogramming code to do that manually, the author chose to use the existing code in std::tuple. std::tuple_element extracts the type of the nth element of a tuple.
static const std::size_t arity = sizeof...(Arguments);
sizeof... 用于获取可变参数模板参数包的大小，因此这一行将函数的参数个数存储在静态整数成员 arity.
sizeof... is used to get the size of a variadic template parameter pack, so this line stores the number of arguments for the function in the static integral member arity.
这篇关于了解函数特征模板的工作原理.特别是成员函数的指针是怎么处理的的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！