为什么匿名命名空间中的函数/对象有外部链接？ [英] Why do functions/objects inside anonymous namespace have external linkage?

问题描述

为什么在匿名命名空间中定义的符号（函数和变量）不具有static关键字的内部链接？

Why don't symbols (functions and variables) that are defined in an anonymous namespace have internal linkage as with static keyword? If a function is not visible/accessible outside, what is the reason to have external linkage?

推荐答案

在C ++ 03中，如果一个函数在外部不可见/具有内部链接的名称被禁止用作模板参数[*]。因此，未命名的命名空间中的大多数事物的名称具有外部链接，以允许它们与模板一起使用。您可以通过声明 static 在命名或全局命名空间中，在未命名的命名空间中显式地给出一个名称内部链接。

In C++03, names with internal linkage were forbidden from being used as template arguments[*]. So, names of most things in unnamed namespaces had external linkage to allow their use with templates. You could explicitly give a name internal linkage in an unnamed namespace by declaring it static, same as in a named or global namespace.

C ++ 11中的改变 - 未命名命名空间中的名称默认为内部链接（3.5 / 4），内部链接的名称可以用作模板参数。

Both things changed in C++11 -- names in unnamed namespaces have internal linkage by default (3.5/4), and names with internal linkage can be used as template arguments.

[*]类型，它必须有外部链接。对于对象和函数，如果其地址用作模板参数，则它必须具有外部链接，但是例如可以使用const的值作为模板参数带内部链接的整数。

[*] for types, it must have external linkage. For objects and functions, it must have external linkage if its address is used as a template argument, although it's OK for example to use as a template argument the value of a const integer with internal linkage.

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