基于CRTP的解决方案是什么样子的？ [英] What would a CRTP-based solution to this look like?

问题描述

我在此信息中提出了以下问题（为方便起见，请在下面粘贴）。其中一个意见认为，有一个基于CRTP的解决方案。我不能弄清楚CRTP是如何相关的（好吧，我从来没有使用过CRTP，所以我不习惯这些术语）。

I asked the following question in this post (pasted below for convenience). One of the comments suggested that there is a CRTP-based solution to the problem. I am not able to figure out how CRTP is relevant here (well, I never used CRTP before, so I am not used to thinking in these terms). So, how would a CRTP-based solution look like?

这是引用的问题：

它是一个基于CRTP的解决方案可能写一个模板函数，它将拥有模板参数的基类的类型信息？ （假设模板参数只来自一个类）

Is it possible to write a template function that would possess type information about the base class of the template argument? (assuming that the template argument derives from one class only)

所以，我在寻找这样的东西：

So, I am looking for something like this:

template <class T> 
auto f(T t) -> decltype(...) { // ... is some SFINAE magic that 
                               //     catches B, the base of T
    std::cout << (B)t << std::endl;
}  

一些相关背景：实现 A * 算法。模板参数是 Node 结构。因此，用户可以定义：

Some relevant background: I am writing a generic implementation of the A* algorithm. The template argument is a Node structure. So, the user might define:

struct NodeBase {
    REFLECTABLE((double)g, (double)f)
        // Using the REFLECTABLE macro as described here:                    
        // http://stackoverflow.com/a/11744832/2725810 
};

struct NodeData : public NodeBase {
    using Base1 = NodeBase;
    REFLECTABLE((double)F)
};

我想编写一个打印节点结构内容的函数。 REFLECTABLE 执行提取 struct 的字段的所有艰苦工作。但是，当用户给我一个 NodeData 实例时，我的函数需要打印 NodeBase 组件的内容。我想稍后为两个和三个基类添加我的函数的重载。

I would like to write a function that prints the contents of the node structure. REFLECTABLE does all the hard work of extracting the fields of the struct. However, when the user gives me a NodeData instance, my function needs to print the contents of the NodeBase component as well. I would like to later add overloads of my function for two and three base classes.

推荐答案

基类我们有std :: is_base_of<>模板结构，它可以与部分专门化或std :: enable_if结合使用。

to know whether a class derives from a base class we have the std::is_base_of<> template structure, which can be used in conjunction with partial specialisation, or std::enable_if.

这是一个演示使用部分专用的结构来应用一个操作取决于它是从node_base派生的（在这种情况下，它只打印基础对象，但你可以做任何其他操作）

Here is a demonstration of using a partially specialised structure to apply a an operation depending on whether it's derived from node_base or not (in this case, it just prints the base object but you could do any other operation)

#include <iostream>
#include <type_traits>

// base class
struct node_base
{

};

std::ostream& operator<<(std::ostream& os, const node_base& nb)
{
    os << "node_base_stuff";
    return os;
}

// a class derived from node_base
struct node : public node_base
{

};

// a class not derived from node_base    
struct not_node
{

};

// apply the general case - do nothing
template<class T, class = void>
struct report_impl
{
    static void apply(const T&) {};
};

// apply the case where an object T is derived from node_base    
template<class T>
struct report_impl<T, std::enable_if_t< std::is_base_of<node_base, T>::value > >
{
    static void apply(const T& t) {
        std::cout << static_cast<const node_base&>(t) << std::endl;
    };
};

// the general form of the report function defers to the partially
// specialised application class
template<class T>
void report(const T& t)
{
    report_impl<T>::apply(t);
}

using namespace std;

// a quick test    
auto main() -> int
{
    node n;
    not_node nn;
    report(n);
    report(nn);

    return 0;
}

预期输出：

node_base_stuff

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