使用模板元编程计数？ [英] Counting With Template Metaprogramming?

问题描述

我一直试图想出一个创造性的解决方案（开启和关闭）一段时间，但我还没有能够。我最近认为它可能是模板元编程可以解决的，虽然我不确定，因为我相对缺乏该技术的经验。

I've been trying to think up a creative solution to this problem (on and off) for some time, but I have not as of yet been able to. I recently considered that it might be solvable with template metaprogramming, though I am not sure due to my relative lack of experience with the technique.

是否可以使用模板元编程（或使用C ++语言的任何其他机制）来计算从某个基类派生的类的数量，以便为每个派生类赋予唯一的静态类标识符？

Is it possible to use template metaprogramming (or any other mechanism with the C++ language) to count the number of classes which are derived from some base class such that each derived class is given a unique, static class identifier?

提前致谢！

推荐答案

没有。这是一个在实践中出现的问题，据我所知，只有两个解决方案：

No. This is a problem that comes up in practice quite a lot, and as far as I'm aware there are only two solutions:

1. 手动为每个派生类分配ID。


2. 动态且懒惰地生成ID。但
   

你做第二个的方式是这样的：

The way you do the second one is something like this:

class Base
{
    virtual int getId() const = 0;
};

// Returns 0, 1, 2 etc. on each successive call.
static int makeUniqueId()
{
    static int id = 0;
    return id++;
}

template <typename Derived>
class BaseWithId : public Base
{
    static int getStaticId()
    {
        static int id = makeUniqueId();
        return id;
    }

    int getId() const { return getStaticId(); }
};

class Derived1 : public BaseWithId<Derived1> { ... };
class Derived2 : public BaseWithId<Derived2> { ... };
class Derived3 : public BaseWithId<Derived3> { ... };

这为您提供每个班级的唯一ID：

This gives you unique IDs for each class:

Derived1::getStaticId(); // 0
Derived2::getStaticId(); // 1
Derived3::getStaticId(); // 2

但是，这些ID是懒惰分配的，所以您调用的订单 getId（）会影响返回的ID。

However, those IDs are assigned lazily, so the order you call getId() affects the ID's returned.

Derived3::getStaticId(); // 0
Derived2::getStaticId(); // 1
Derived1::getStaticId(); // 2

您的应用程序是否合适取决于您的特定需求（例如，没有序列化的好处。）

Whether or not this is OK for your application depends on your particular needs (e.g. would be no good for serialisation).

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