模板多态性

2022-01-24 00:00:00 templates polymorphism c++

我有这种类结构.

class Interface {
  // ...
};

class Foo : public Interface {
  // ...
};

template <class T>
class Container {
  // ...
};

我还有其他一些 Bar 类的构造函数.

And I have this constructor of some other class Bar.

Bar(const Container<Interface> & bar){
  // ...
}

当我以这种方式调用构造函数时，我得到一个没有匹配的函数"；错误.

When I call the constructor this way I get a "no matching function" error.

Container<Foo> container ();

Bar * temp = new Bar(container);

怎么了?模板不是多态的吗?

What is wrong? Are templates not polymorphic?

推荐答案

我认为你需要的确切术语是模板协方差"，这意味着如果 B 继承自 A，那么 T<B> 继承自 T<A>.在 C++ 中不是这种情况，Java 和 C# 泛型*也不是.

I think the exact terminology for what you need is "template covariance", meaning that if B inherits from A, then somehow T<B> inherits from T<A>. This is not the case in C++, nor it is with Java and C# generics*.

有一个很好的理由来避免模板协变:这将简单地删除模板类中的所有类型安全.让我用下面的例子来解释:

There is a good reason to avoid template covariance: this will simply remove all type safety in the template class. Let me explain with the following example:

//Assume the following class hierarchy
class Fruit {...};

class Apple : public Fruit {...};

class Orange : public Fruit {...};

//Now I will use these types to instantiate a class template, namely std::vector
int main()
{
    std::vector<Apple> apple_vec;
    apple_vec.push_back(Apple()); //no problem here

    //If templates were covariant, the following would be legal
    std::vector<Fruit> & fruit_vec = apple_vec;

    //push_back would expect a Fruit, so I could pass it an Orange
    fruit_vec.push_back(Orange()); 

    //Oh no! I just added an orange in my apple basket!
}

因此，无论 A 和 B 之间的关系如何，您都应该将 T<A> 和 T<B> 视为完全不相关的类型.

Consequently, you should consider T<A> and T<B> as completely unrelated types, regardless of the relation between A and B.

那么您如何解决您面临的问题?在 Java 和 C# 中，您可以分别使用 有界通配符 和 约束:

So how could you solve the issue you're facing? In Java and C#, you could use respectively bounded wildcards and constraints:

//Java code
Bar(Container<? extends Interface) {...}

//C# code
Bar<T>(Container<T> container) where T : Interface {...}

下一个 C++ 标准(称为 C++1x(以前称为 C++0x))最初包含一个更强大的机制，名为 概念，这将使开发人员对模板参数强制执行语法和/或语义要求，但不幸的是推迟到以后的日期.但是，Boost 有一个可能感兴趣的 Concept Check 库你.

The next C++ Standard (known as C++1x (formerly C++0x)) initially contained an even more powerful mechanism named Concepts, that would have let developers enforce syntaxic and/or semantic requirements on template parameters, but was unfortunately postponed to a later date. However, Boost has a Concept Check library that may interest you.

尽管如此，对于您遇到的问题，概念可能有点矫枉过正，使用 @gf 可能是最好的解决方案.

Nevertheless, concepts might be a little overkill for the problem you encounter, an using a simple static assert as proposed by @gf is probably the best solution.

* 更新:从 .Net Framework 4 开始，可以将泛型参数标记为 协变或逆变.

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