否，它无效，但是在您的示例中是有效的，因为最后一个 const 实际上是签名(隐藏的 Foo * this 第一个参数现在为 const Foo * this ).

no, it is not valid, but in your example it is, because the last const is actually part of the signature (the hidden Foo *this first parameter is now const Foo *this).

它用于以只读方式访问(获取const引用，方法是常量)，或进行写入(获取非const引用，方法不是常量)

It is used to access in read-only (get const reference, the method is constant), or write (get non-const reference, the method is not constant)

在两种方法中都返回相同实体(常量或非常量)的引用仍然是一个不错的设计选择！

it's still a good design choice to return the reference of the same entity (constant or non-constant) in both methods of course!