在 using 语句中指定 Class 名称有什么作用？

Question

鉴于以下情况：

namespace Foo{
class Bar{
    static const auto PRIVATE = 0;
    const int private_ = 1;
    void ptivateFunc() { cout << 2; }
public:
    static const auto PUBLIC = 3;
    const int public_ = 4;
    void publicFunc() { cout << 5; }
};
}

语句 using Foo::Bar; 编译...但我不确定它为我提供了访问权限。任何人都可以解释该声明的重点是什么以及它可以让我获得关于 Bar 的内容，而不是简单地做一个 using namespace Bar?

Answer 1

它允许您在没有 Foo 命名空间的情况下使用 Bar。

Answer 2

来自 cppreference:

using ns_name::name; (6)
(...)
6) using-declaration: makes the symbol name from the namespace ns_name accessible for unqualified lookup as if declared in the same class scope, block scope, or namespace as where this using-declaration appears.

using namespace ns_name; (5)
5) using-directive: From the point of view of unqualified name lookup of any name after a using-directive and until the end of the scope in which it appears, every name from namespace-name is visible as if it were declared in the nearest enclosing namespace which contains both the using-directive and namespace-name.

所以基本上你可以在命名空间 Foo 之外（但在 using 声明的范围内）写 Bar 而不是 Foo::Bar，而来自命名空间 Foo 还需要全名。

如果您使用 using namespace Foo，您可以通过本地名称访问 Foo 中的所有符号，而无需显式 Foo::.

在 using 语句中指定 Class 名称有什么作用？

What Does Specifying a Class Name in a using Statement Do?

c++

scope

named-scope

using

class