使用可变参数模板的 mixin 继承的可见性规则

Question

假设我从可变参数模板继承了整个参数列表。参数是如何继承的？

// snippet
template<typename... R>
class foo
    : public R... {
public: 
};
// ....
using foo_inst = foo<bar_1, bar_2>;

我试过了，好像所有 R的都是继承的public（不只是第一个）。 这是定义的行为吗？

我用 gcc 和 msvc 试过了（感谢 also with clang), all with same results. The compilers not even mentioned any warnings. You can see a running example here。

Answer 1

是的，这是定义的行为。引用自 14.5.3[temp.variadic]

A pack expansion consists of a pattern and an ellipsis, the instantiation of which produces zero or more instantiations of the pattern in a list (described below). The form of the pattern depends on the context in which the expansion occurs. Pack expansions can occur in the following contexts:

该列表中的相关上下文是：

— In a base-specifier-list (Clause 10 ); the pattern is a base-specifier.

因此，在参数包扩展class foo : public R...中，模式是base-specifierpublic R，使得包由类型[=13]组成=] 扩展为 public T1, public T2, ... , public Tn。（上一句中的省略号用于数学序列指定意义。）

使用可变参数模板的 mixin 继承的可见性规则

Visibility rules for mixin inheritance with variadic template

c++

inheritance

template-mixins

variadic-templates

c++11