C ++：指定模板参数的基类 [英] C++: specifying a base class for a template parameter

问题描述

我需要设计一个框架，并行计算分治算法的结果。
为了使用框架，用户需要指定实现分割阶段（从T到T的函数），征服阶段（从D到D的函数）和T和D本身。

I need to design a framework that computes the result of a divide-et-conquer algorithm in parallel. In order to use the framework, the user needs to specify somehow the procedure that implements the "divide" phase (a function from T to T), the "conquer" phase (a function from D to D) and T and D themselves.

我认为定义两个抽象类， BaseDivide 和 BaseConquer ，它用正确的类型声明一个纯虚方法 compute ：这样我有一个类型，概念（从框架的角度），通过派生抽象类包含用户可定义的函数。

I've thought it would be nice to define two abstract classes, BaseDivide and BaseConquer, which declares a pure virtual method compute with the right types: that way I have a type which implements a well-defined concept (from the point of view of the framework) with the user-definable function included by means of derivation of the abstract classes.

我曾经想过使用模板传递类型到框架，所以用户不必实例化它们为了使用框架，所以像这样：

I've thought to use templates to pass the types to the framework, so the user doesn't have to instantiate them in order to use the framework, so something like that:

template <typename T, typename D, typename Divide, typename Conquer> 
D compute(T arg);

我的问题是我想要Divide and Conquer是 BaseDivide 和 BaseConquer ：有一种方法在编译时强制执行它？

My problem is that I want that Divide and Conquer to be derived types of BaseDivide and BaseConquer: there is a way to enforce it at compile time? Also: do you think I can achieve a similar result with a cleaner design?

推荐答案

您可以创建如下的基类：

You could create the base classes like this:

struct BaseDivide {
    enum EnumDiv { derivedFromBaseDivide = true };
}

template <typename T, typename D, typename Divide, typename Conquer> 
    static_assert(D::derivedFromBaseDivide);
    D compute(T arg);

额外的Divide and Conquer模板参数的用途是什么？您确定需要他们吗？

What is the purpose of the additional Divide and Conquer template parameters? Are you sure you need them?

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