类模板可以实例化没有成员？ [英] Class template can be instantiated without members?

问题描述

维基百科文章说明了这一点：

实例化类模板不会导致其成员定义被实例化。

instantiating a class template does not cause its member definitions to be instantiated.

我无法想象任何类

推荐答案

许多早期的C ++编译器被实例化

Many early C++ compilers instantiated all member functions, whether you ever called them or not.

例如，考虑 std :: list ，它有一个 sort 成员函数。使用当前正常运行的编译器，您可以在不支持比较的类型上实例化 list 。如果您尝试使用 list :: sort ，它将失败，因为您不支持比较。只要你不为该列表调用 sort ，就可以了，因为 list< T> :: sort

Consider, for example, std::list, which has a sort member function. With a current, properly functioning compiler, you can instantiate list over a type that doesn't support comparison. If you try to use list::sort, it will fail, because you don't support comparison. As long as you don't call sort for that list, it's all fine though, because list<T>::sort won't be instantiated unless you call it.

然而，对于那些老的，运行不良的编译器，试图创建 list< T> 意味着 list< T> :: sort 已实例化，即使您从未使用它。存在 list :: sort 意味着您需要为 T / code>，即使您从未实际使用过 sort ，也只是为了创建列表< T&该类型的列表。

With those older, poorly functioning compilers, however, trying to create list<T> meant that list<T>::sort was instantiated even though you never used it. The existence of list::sort meant that you needed to implement < for T, just to create a list<T>, even if you never actually used sort on a list of that type at all.

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