How does raft handle committing entries from previous one?

北战南征 提交于 2019-12-04 12:06:15

问题


In raft paper section 5.4.2

If a leader crashes before committing an entry, future leaders will attempt to finish replicating the entry. However, a leader cannot immediately conclude that an entry from a previous term is committed once it is stored on a majority of servers. There could be a situation where an old log entry is stored on a majority of servers, yet can still be overwritten by a future leader.

The author mentioned that to avoid the situation above

To eliminate problems like the one in Figure 8, Raft never commits log entries from previous terms by counting replicas. Only log entries from the leader’s current term are committed by counting replicas; once an entry from the current term has been committed in this way, then all prior entries are committed indirectly because of the Log Matching Property.

But wouldn't the same problem still occur?

Given the following situation that the author provided

When S5 is elected leader it only looks at its current committed log which is (term3, index1) and this is going to override term2 entries in all followers.

How does making a leader looking at its own committed log solve the problem?


回答1:


Read the caption on this image. Both (d) and (e) are possible resolutions to the state of the log produced by (a), (b), and (c). The problem is, even though in (c) entry (2, 2) was replicated to a majority of the cluster, this is illustrating that it could still be overwritten when S5 is elected in (d). So, the solution is to only allow nodes to commit entries from their own term. In other words, replicating an entry on a majority of nodes does not equal commitment. In (c), entry (2, 2) is replicated to a majority of the cluster, but because it's not in the leader's term (at least 4) it's not committed. But in (e), after the leader replicates an entry from its current term (4), that prevents the situation in (d) from occurring since S5 can no longer be elected leader.




回答2:


After S1 replicates entry 4 with a higher term than 2 and 3. S5 will no longer be elected as leader, since the leader election strategy of Raft:

Raft determines which of two logs is more up-to-date by comparing the index and term of the last entries in the logs. If the logs have last entries with different terms, then the log with the later term is more up-to-date. If the logs end with the same term, then whichever log is longer is more up-to-date.

So, in my opinion, the appended log entry 4 in (e) implicitly promote all the entries' term before it. Because what we only care about is the term or the last entry, rather than entry 2 any more.

This is just like what the proposer do in Phase 2 of Paxos:

If the proposer receives a response to its prepare requests (numbered n) from a majority of acceptors, then it sends an accept request to each of those acceptors for a proposal numbered n with a value v, where v is the value of the highest-numbered proposal among the responses, or is any value if the responses reported no proposals.

That's say, propose the learned value 2 with a higher propose number.




回答3:


I think both situations in figure 8 (d) and (e) are legal in Raft because the paper says:

To eliminate problems like the one in Figure 8, Raft never commits log entries from previous terms by counting replicas. Only log entries from the leader’s current term are committed by counting replicas.

In figure 8(d) the entries with term 2 is not in the local log of leader S5, and they are not committed to the state machine. It is ok to overwrite them with entries with term 3. Only entries in leader's current log are eligible to be considered as committed by counting the number of replicas.




回答4:


If we allow entry from the previous term being committed, after (c), entry numbered 2 will be committed. After that, if 3 is selected as the leader, it will overwrite the committed 2. Thus, S5 and S1 will execute different commands. To prevent that, we will not allow 2 committed. Thus, the commands that are executed in all state machines will become consistent.



来源:https://stackoverflow.com/questions/46223417/how-does-raft-handle-committing-entries-from-previous-one

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