内部模板类型 std::vector<std::vector<T>> 的函数模板重载或特化 [英] Function Template Overloading or Specialization for inner template type std::vector<std::vector<T>>

问题描述

如何实现内部模板类型std::vector>的模板重载功能.

How to achieve function Template Overloading for inner template type std::vector<std::vector<T>>.

我有一个重载模板程序和一个包含映射、对和向量的复杂数据结构.

I have a program of overloaded templates and a complex data structure having maps, pairs and vectors.

#include <iostream>
#include <vector>
#include <map>
#include <utility>
#include <typeinfo>

template<typename Test, template<typename...> class Ref> //#6
struct is_specialization : std::false_type {};

template<template<typename...> class Ref, typename... Args> //#7
struct is_specialization<Ref<Args...>, Ref>: std::true_type {};
template <typename T>
bool f(T& x) // #1
{
    std::cout << "body of f\n";
    return f(x);
}

template <typename T>
bool f(std::vector<T>& v) // #2
{
    std::cout << "body of f for vectors\n";
    return true;
}
template<typename T>
typename std::enable_if<is_specialization<typename T::value, std::vector>::value, T>::type
    bool f(std::vector<T>& v) // #5
    {
        std::cout << "body of f for vectors<vectors>\n";
        return true;
    }

template <typename Key, typename Value>
bool f(const std::pair<Key,Value>& v) // #3
{
    std::cout << "body of f for pairs\n";
    for(auto& e: v) {
      f(e.first);
    }
    for(auto& e: v) {
      f(e.second);
    }
    return true;
}

template <typename Key, typename Value>
bool f(std::map<Key,Value>& v) // #4
{
    std::cout << "body of f for maps\n";
    for(auto& e: v) {
      f(e.first);  // expecting this call goes to #3
    }
    for(auto& e: v) {
      f(e.second);
    }
    return true;
}

int main() {
  std::vector<int> v{1,2};
  std::map<std::pair<int,int>,std::vector<std::vector<int>>> m_map = {
                                            {{10,20}, {{5,6},{5,6,7}}},
                                            {{11,22}, {{7,8},{7,8,9}}}
                                        };
    f(m_map); // this call goes to #4
} 

总是调用向量 #2，但是对于 std::vectors<std::vector<T>> 我需要调用 #5 并且我收到编译错误 wrtstd::enable_if 中使用的 ::type.请让我知道这个程序有什么问题以及如何使它工作.也有人可以解释一下#6 和#7 表示 w.r.t 模板参数包是什么，它是如何工作的.

Always for vectors #2 is getting called, but for std::vectors<std::vector<T>> I need #5 to get called and also I am getting compilation error w.r.t ::type used in std::enable_if. Please let me know what is wrong in this program and how to make it work. Also Can some one explain what does #6 and #7 signify w.r.t template parameter pack, how does it work.

谢谢.

推荐答案

我看到的为 std::vector<std::vector<T>> 专门化编写的最简单方法>f() 如下

The simplest way I see to write a std::vector<std::vector<T>> specialization for f() is the following

template<typename T>
bool f (std::vector<std::vector<T>>& v) // #5
 {
   std::cout << "body of f for vectors<vectors>\n";
   return true;
 }

这样你就重载了比 #2 更专业的模板 f() 函数.

This way you have overloaded template f() function that is more specialized than #2.

如果您想将 SFINAE 与您的 is_specialization 一起使用，在我看来正确的方法如下

If you want to use SFINAE with your is_specialization, seems to me that the right way is the following

template <typename T>
typename std::enable_if<is_specialization<T, std::vector>::value, bool>::type
f (std::vector<T> & v) // #5
 {
   std::cout << "body of f for vectors<vectors>\n";
   return true;
 }

不幸的是，这个版本专门作为版本 #2，所以当你用 std::vector 调用 f() 时;>，你会得到一个歧义，所以编译错误.

Unfortunately this version is specialized as version #2, so when you call f() with a std::vector<std::vector<T>>, you get an ambiguity so a compilation error.

要解决此问题，您还必须禁用#2 版本

To solve this problem you have also to disable the #2 version

template <typename T>
typename std::enable_if<! is_specialization<T, std::vector>::value, bool>::type
f (std::vector<T> & v) // #2
 {
   std::cout << "body of f for vectors\n";
   return true;
 }

在您的原始版本中...您使用 typename T::type...但是当 T 不是带有 type 已定义.

In your original version... you use typename T::type... but this gives an error when T isn't a class with a type defined.

更多:返回两种类型

template<typename T>
typename std::enable_if<is_specialization<typename T::value,
                        std::vector>::value, T>::type // <<--- type 1: T
    bool f(std::vector<T>& v) // #5
//  ^^^^  type2: bool

这样使用SFINAE，返回的类型必须用std::enable_if

Using SFINAE this way, the returned type has to be expressed by std::enable_if

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