问题描述
如果我有一组连接的对等体，让我们说3（a-> b，a-> c，a-> d）
我只想断开对等体c，我该怎么办？在做什么？
If I have an array of connected peers, let's say 3 (a->b, a->c, a->d)and I want to disconnect peer "c" only, what should I be doing?
我看到一个对类似问题的回复表明你只能断开自己与会话的关系：这意味着在上面的场景中，如果我这样做：
I've seen one response to a similar question state that you can only disconnect 'yourself' from a session: meaning that in the above scenario if I do this:
[mySession disconnect];
我同时从b，c和d同时断开a 。
所以建议是向你想要断开连接的对等体发送通知（c）并告诉它断开与会话的连接。
that I all be disconnecting "a" from "b", "c" and "d" simultaneously.So the suggestion was to send a notification to the peer you want disconnected ("c") and tell it to disconnect from the session.
然而，我我也看到了这个，并想知道它是否符合我的要求 - 仅删除特定的同行：
However, I've also seen this and wondered if it does what I'm asking - removes the specific peer only:
[mySession.connectedPeers[0] disconnect];
编辑：我刚刚意识到这是一个只读属性，所以忽略这一部分。
I just realized this is a read-only property, so ignore this portion.
或者还有另一种更好的方法可以删除特定的同行吗？
Or is there yet another, perhaps better way to remove specific peers?
推荐答案
每个对等连接由MCSession的实例表示。
因此在您的示例中，您将有3个指向MCSession对象的UNIQUE指针：
Each peer-to-peer connection is represented by an instance of MCSession.So in your example, you will have 3 UNIQUE pointers to MCSession objects:
MCSession *p1 = a->b
MCSession *p2 = a->c
MCSession *p3 = a->d
所以要断开对等c，你将：
so to disconnect peer c you will:
[p2 disconnect]; // release the session
p2 = nil; // release the resource
当然，您可以将会话存储在可变数组中，然后执行相同操作逻辑，但使用数组中的指针
Of course, you can store sessions in a mutable array and then do the same logic, but use the pointer from the array
p2 = (MCSession *)[array objectAtIndex:1];
[p2 disconnect];
[array removeObjectAtIndex:1];
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