Is a class (or struct) that implements a more-derived interface also required to state in its base type list that it is implementing all of those less-derived interfaces?
一个实现了派生程度较高的接口的class或者struct必须要在其基类型列表中声明它实现了那些派生程度较低的接口吗？
So a class or struct is required to implement all the methods of all the interfaces in the transitive closure of its base interfaces?
那么就是说一个class或者一个struct必须要实现其传递闭包上所有接口定义的所有方法吗？
A struct can implement interfaces.
结构可以实现接口。
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