如何扩展对可变参数模板基类的调用?

时间:2023-03-10
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问题描述

我有一组非正交策略,它们都实现了一个通用的命名方法,这些策略添加了安全检查.我希望用户能够组合策略以允许更复杂的验证,而无需手动为每个组合案例创建策略.我的方法是创建一个新的策略类来组合其他类.

I have a set of non-orthogonal policies, all of them implementing a common named method, the policies add safety checks. I want users to be able to combine the policies to allow more complex validation without creating policies for each combination case by hand. My approach is creating a new policy class to combine others.

下面的简化示例将 C 显示为组合类,这里组合了方法 id.预期的结果是,在 C 上调用 id 时,依次调用每个基类的 id.

The simplified example below shows C as the combining class, here the method id is combined. The expected result is, when calling id on C, to sequentially call the id of each base class.

#include <iostream>
using namespace std;

struct A 
{
    void id() { cout << "A ";}
};

struct B 
{
    void id() { cout << "B ";}
};

template<class A, class... As>
struct C : public A, public As... 
{
    void id()
    {
         A::id();
         As...::id(); // This line does not work, it is illustrative.
    }
};

int main()
{
    C<A, B> c;
    c.id();

    //expected: result A B 
}

问题是:是否可以扩展 As... 以某种方式在不使用递归方法的情况下执行此操作,仅使用 ... 运算符?

The question is: Is it possible to expand As... somehow to do this without using a recursive approach, just using the ... operator?

推荐答案

好的.您需要一个允许包扩展的上下文——一个简单的上下文是一个带括号的初始化列表,它也有保证从左到右求值的好处:

Sure. You need a context that permits pack expansion - a simple one is a braced initializer list, which also has the benefit of guaranteeing left-to-right evaluation:

using expander = int[];
(void) expander { 0, ((void) As::id(), 0)... };

  • ... 向左扩展一个模式;在这种情况下,模式是表达式 ((void) As::id(), 0).

    • ... expands a pattern to its left; in this case the pattern is the expression ((void) As::id(), 0).

      表达式中的,是逗号运算符,它计算第一个操作数,丢弃结果,然后计算第二个操作数,并返回结果.

      The , in the expression is the comma operator, which evaluates the first operand, discards the result, then evaluates the second operand, and returns the result.

      演示.

      在 C++17 中(如果幸运的话),C::id 的整个主体可以替换为 一个二元折叠表达式:(A::id(), ... , (void) As::id()); 演示.

      In C++17 (if we are lucky), the entire body of C::id can be replaced with a binary fold expression: (A::id(), ... , (void) As::id()); Demo.

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