如何正确声明自引用模板类型? [英] How to properly declare a self-referencing template type?

问题描述

如何声明引用自身的模板类型?

 模板< class T = Animal>动物类{上市:T getChild();} 

这样，我得到一个有关缺少类型说明符的编译器错误.我试图向前声明 Animal ，但没有成功.

我试图强加类型约束. Lion 只能有一个 Lion 作为孩子，一个 Bear 有一个 Bear ，依此类推./p>

编辑

我将发布实际课程的一部分.它是可以出现在链接列表中的类的模板:

 模板< class T =链接的< T>>类链接{私人的:T * m_prev;T * m_next;} 

我要强制该类只能指向同一类(或子类)的对象.

解决方案

做类似链表的常用方法是:

 模板< class T>类链接{私人的:链接的T * m_prev;链接的T * m_next;} 

这对您有用吗?如果没有，那么您试图完成的工作无法通过这种方式完成吗?

How do I declare a templated type that refers to itself?

template <class T = Animal> class Animal
{
public:
    T getChild ();
}

With this, I get a compiler error concerning a missing type specifier. I tried to forward-declare Animal, without success.

I am trying to impose a type constraint. A Lion can only have a Lion as a child, a Bear has a Bear, and so on.

EDIT

I'll post part of the actual class. It is a template for classes that can appear in a linked list:

template <class T = Linked<T> > class Linked
{
private:
    T* m_prev;
    T* m_next;
}

I want to enforce that the class can only point to object of the same class (or a subclass).

解决方案

The usual way to do something like a linked list is:

template <class T> class Linked
{
private:
    Linked<T>* m_prev;
    Linked<T>* m_next;
}

Does this work for you, and if not, what are you trying to accomplish that can't be done this way?

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