golang[113]-raft理论与实践[4]-lab2b日志复制

准备工作

1、阅读raft论文
2、阅读raft理论与实践[1]-理论篇
3、阅读raft理论与实践[2]-lab2a
4、阅读raft理论与实践[3]-lab2a讲解
5、查看我写的这篇文章: 模拟RPC远程过程调用

执行日志

我们需要执行日志中的命令,因此在make函数中,新开一个协程:applyLogEntryDaemon()

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func Make(peers []*labrpc.ClientEnd, me int,
persister *Persister, applyCh chan ApplyMsg) *Raft {
...
go rf.applyLogEntryDaemon() // start apply log
DPrintf("[%d-%s]: newborn election(%s) heartbeat(%s) term(%d) voted(%d)\n",
rf.me, rf, rf.electionTimeout, rf.heartbeatInterval, rf.CurrentTerm, rf.VotedFor)
return rf
}

一个死循环
1、如果rf.lastApplied == rf.commitIndex, 意味着commit log entry命令都已经被执行了,这时用信号量陷入等待。
一旦收到信号,说明需要执行命令。这时会把最后执行的log entry之后,一直到最后一个commit log entry的所有log都传入通道apply中进行执行。
由于是测试,处理apply的逻辑会在测试代码中。

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// applyLogEntryDaemon exit when shutdown channel is closed
func (rf *Raft) applyLogEntryDaemon() {
for {
var logs []LogEntry
// wait
rf.mu.Lock()
for rf.lastApplied == rf.commitIndex {
rf.commitCond.Wait()
select {
case <-rf.shutdownCh:
rf.mu.Unlock()
DPrintf("[%d-%s]: peer %d is shutting down apply log entry to client daemon.\n", rf.me, rf, rf.me)
close(rf.applyCh)
return
default:
}
}
last, cur := rf.lastApplied, rf.commitIndex
if last < cur {
rf.lastApplied = rf.commitIndex
logs = make([]LogEntry, cur-last)
copy(logs, rf.Logs[last+1:cur+1])
}
rf.mu.Unlock()
for i := 0; i < cur-last; i++ {
// current command is replicated, ignore nil command
reply := ApplyMsg{
CommandIndex: last + i + 1,
Command: logs[i].Command,
CommandValid: true,
}
// reply to outer service
// DPrintf("[%d-%s]: peer %d apply %v to client.\n", rf.me, rf, rf.me)
DPrintf("[%d-%s]: peer %d apply to client.\n", rf.me, rf, rf.me)
// Note: must in the same goroutine, or may result in out of order apply
rf.applyCh <- reply
}
}
}

新增 Start函数,此函数为leader执行从client发送过来的命令。
当client发送过来之后,首先需要做的就是新增entry 到leader的log中。并且将自身的nextIndex 与matchIndex 更新。

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func (rf *Raft) Start(command interface{}) (int, int, bool) {
index := -1
term := 0
isLeader := false

// Your code here (2B).
select {
case <-rf.shutdownCh:
return -1, 0, false
default:
rf.mu.Lock()
defer rf.mu.Unlock()
// Your code here (2B).
if rf.state == Leader {
log := LogEntry{rf.CurrentTerm, command}
rf.Logs = append(rf.Logs, log)

index = len(rf.Logs) - 1
term = rf.CurrentTerm
isLeader = true

//DPrintf("[%d-%s]: client add new entry (%d-%v), logs: %v\n", rf.me, rf, index, command, rf.logs)
DPrintf("[%d-%s]: client add new entry (%d)\n", rf.me, rf, index)
//DPrintf("[%d-%s]: client add new entry (%d-%v)\n", rf.me, rf, index, command)

// only update leader
rf.nextIndex[rf.me] = index + 1
rf.matchIndex[rf.me] = index
}
}

return index, term, isLeader
}

接下来最重要的部分涉及到日志复制,这是通过AppendEntries实现的。我们知道leader会不时的调用consistencyCheck(n)进行一致性检查。
在给第n号节点一致性检查时,首先获取pre = rf.nextIndex,pre至少要为1。代表要给n节点发送的log index。因此AppendEntriesArgs参数中,PrevLogIndex 与 prevlogTerm 都为pre - 1位置。
代表leader相信PrevLogIndex及其之前的节点都是与leader相同的。
将pre及其之后的entry 加入到AppendEntriesArgs参数中。 这些log entry可能是与leader不相同的,或者是follower根本就没有的。

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func (rf *Raft) consistencyCheck(n int) {
rf.mu.Lock()
defer rf.mu.Unlock()
pre := max(1,rf.nextIndex[n])
var args = AppendEntriesArgs{
Term: rf.CurrentTerm,
LeaderID: rf.me,
PrevLogIndex: pre - 1,
PrevLogTerm: rf.Logs[pre - 1].Term,
Entries: nil,
LeaderCommit: rf.commitIndex,
}

if rf.nextIndex[n] < len(rf.Logs){
args.Entries = append(args.Entries, rf.Logs[pre:]...)
}

go func() {
DPrintf("[%d-%s]: consistency Check to peer %d.\n", rf.me, rf, n)
var reply AppendEntriesReply
if rf.sendAppendEntries(n, &args, &reply) {
rf.consistencyCheckReplyHandler(n, &reply)
}
}()
}

接下来查看follower执行AppendEntries时的反应。
AppendEntries会新增两个返回参数:
ConflictTerm代表可能发生冲突的term
FirstIndex 代表可能发生冲突的第一个index。

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type AppendEntriesReply struct {
CurrentTerm int // currentTerm, for leader to update itself
Success bool // true if follower contained entry matching prevLogIndex and prevLogTerm
// extra info for heartbeat from follower
ConflictTerm int // term of the conflicting entry
FirstIndex int // the first index it stores for ConflictTerm
}

如果args.PrevLogIndex < len(rf.Logs), 表明至少当前节点的log长度是合理的。
令preLogIdx 与 args.PrevLogIndex相等。prelogTerm为当前follower节点preLogIdx位置的term。
如果拥有相同的term,说明follower与leader 在preLogIdx之前的log entry都是相同的。因此请求是成功的。
此时会截断follower的log,将传递过来的entry加入到follower的log之后,执行此步骤后,强制要求与leader的log相同了。
请求成功后,reply的ConflictTerm为最后一个log entry的term,reply的FirstIndex为最后一个log entry的index。

否则说明leader与follower的日志是有冲突的,冲突的原因可能是:
1、leader认为的match log entry超出了follower的log个数,或者follower 还没有任何log entry(除了index为0的entry是每一个节点都有的)。
2、log在相同的index下,leader的term 与follower的term确是不同的。
这时找到follower冲突的term即为ConflictTerm。
获取此term的第一个entry的index即为FirstIndex。
所以最后,AppendEntries会返回冲突的term以及第一个可能冲突的index。

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// AppendEntries handler, including heartbeat, must backup quickly
func (rf *Raft) AppendEntries(args *AppendEntriesArgs, reply *AppendEntriesReply) {
...
preLogIdx, preLogTerm := 0, 0
if args.PrevLogIndex < len(rf.Logs) {
preLogIdx = args.PrevLogIndex
preLogTerm = rf.Logs[preLogIdx].Term
}

// last log is match
if preLogIdx == args.PrevLogIndex && preLogTerm == args.PrevLogTerm {
reply.Success = true
// truncate to known match
rf.Logs = rf.Logs[:preLogIdx+1]
rf.Logs = append(rf.Logs, args.Entries...)
var last = len(rf.Logs) - 1

// min(leaderCommit, index of last new entry)
if args.LeaderCommit > rf.commitIndex {
rf.commitIndex = min(args.LeaderCommit, last)
// signal possible update commit index
go func() { rf.commitCond.Broadcast() }()
}
// tell leader to update matched index
reply.ConflictTerm = rf.Logs[last].Term
reply.FirstIndex = last

if len(args.Entries) > 0 {
DPrintf("[%d-%s]: AE success from leader %d (%d cmd @ %d), commit index: l->%d, f->%d.\n",
rf.me, rf, args.LeaderID, len(args.Entries), preLogIdx+1, args.LeaderCommit, rf.commitIndex)
} else {
DPrintf("[%d-%s]: <heartbeat> current logs: %v\n", rf.me, rf, rf.Logs)
}
} else {
reply.Success = false

// extra info for restore missing entries quickly: from original paper and lecture note
// if follower rejects, includes this in reply:
//
// the follower's term in the conflicting entry
// the index of follower's first entry with that term
//
// if leader knows about the conflicting term:
// move nextIndex[i] back to leader's last entry for the conflicting term
// else:
// move nextIndex[i] back to follower's first index
var first = 1
reply.ConflictTerm = preLogTerm
if reply.ConflictTerm == 0 {
// which means leader has more logs or follower has no log at all
first = len(rf.Logs)
reply.ConflictTerm = rf.Logs[first-1].Term
} else {
i := preLogIdx
// term的第一个log entry
for ; i > 0; i-- {
if rf.Logs[i].Term != preLogTerm {
first = i + 1
break
}
}
}
reply.FirstIndex = first
if len(rf.Logs) <= args.PrevLogIndex {
DPrintf("[%d-%s]: AE failed from leader %d, leader has more logs (%d > %d), reply: %d - %d.\n",
rf.me, rf, args.LeaderID, args.PrevLogIndex, len(rf.Logs)-1, reply.ConflictTerm,
reply.FirstIndex)
} else {
DPrintf("[%d-%s]: AE failed from leader %d, pre idx/term mismatch (%d != %d, %d != %d).\n",
rf.me, rf, args.LeaderID, args.PrevLogIndex, preLogIdx, args.PrevLogTerm, preLogTerm)
}
}
}

leader调用AppendEntries后,会执行回调函数consistencyCheckReplyHandler。
如果调用是成功的,那么正常的跟新matchIndex,nextIndex即下一个要发送的index应该为matchIndex + 1。

如果调用失败,说明有冲突。
如果confiicting term等于0,说明了leader认为的match log entry超出了follower的log个数,或者follower 还没有任何log entry(除了index为0的entry是每一个节点都有的)。
此时简单的让nextIndex 为reply.FirstIndex即可。

如果confiicting term不为0,获取leader节点confiicting term 的最后一个log index,此时nextIndex 应该为此index与reply.FirstIndex的最小值。
检查最小值是必须的:
假设
s1: 0-0 1-1 1-2 1-3 1-4 1-5
s2: 0-0 1-1 1-2 1-3 1-4 1-5
s3: 0-0 1-1

此时s1为leader,并一致性检查s3, 从1-5开始检查,此时由于leader有更多的log,因此检查不成功,返回confict term 1, firstindex:2
如果只是获取confiicting term 的最后一个log index,那么nextIndex又是1-5,陷入了死循环。

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func (rf *Raft) consistencyCheckReplyHandler(n int, reply *AppendEntriesReply) {
rf.mu.Lock()
defer rf.mu.Unlock()

if rf.state != Leader {
return
}
if reply.Success {
// RPC and consistency check successful
rf.matchIndex[n] = reply.FirstIndex
rf.nextIndex[n] = rf.matchIndex[n] + 1
rf.updateCommitIndex() // try to update commitIndex
} else {
// found a new leader? turn to follower
if rf.state == Leader && reply.CurrentTerm > rf.CurrentTerm {
rf.turnToFollow()
rf.resetTimer <- struct{}{}
DPrintf("[%d-%s]: leader %d found new term (heartbeat resp from peer %d), turn to follower.",
rf.me, rf, rf.me, n)
return
}

// Does leader know conflicting term?
var know, lastIndex = false, 0
if reply.ConflictTerm != 0 {
for i := len(rf.Logs) - 1; i > 0; i-- {
if rf.Logs[i].Term == reply.ConflictTerm {
know = true
lastIndex = i
DPrintf("[%d-%s]: leader %d have entry %d is the last entry in term %d.",
rf.me, rf, rf.me, i, reply.ConflictTerm)
break
}
}
if know {
rf.nextIndex[n] = min(lastIndex, reply.FirstIndex)
} else {
rf.nextIndex[n] = reply.FirstIndex
}
} else {
rf.nextIndex[n] = reply.FirstIndex
}
rf.nextIndex[n] = min(rf.nextIndex[n], len(rf.Logs))
DPrintf("[%d-%s]: nextIndex for peer %d => %d.\n",
rf.me, rf, n, rf.nextIndex[n])
}
}

当调用AppendEntry成功后,说明follower与leader的log是匹配的。此时leader会找到commited的log并且执行其命令。
这里有一个比较巧妙的方法,对matchIndex排序后取最中间的数。
由于matchIndex代表follower有多少log与leader的log匹配,因此中间的log index意味着其得到了大部分节点的认可。
因此会将此中间的index之前的所有log entry都执行了。
rf.Logs[target].Term == rf.CurrentTerm 是必要的:
这是由于当一个entry出现在大多数节点的log中,并不意味着其一定会成为commit。考虑下面的情况:

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S1: 1 2     1 2 4
S2: 1 2 1 2
S3: 1 --> 1 2
S4: 1 1
S5: 1 1 3

s1在term2成为leader,只有s1,s2添加了entry2.
s5变成了term3的leader,之后s1变为了term4的leader,接着继续发送entry2到s3中。
此时,如果s5再次变为了leader,那么即便没有S1的支持,S5任然变为了leader,并且应用entry3,覆盖掉entry2。
所以一个entry要变为commit,必须:
1、在其term周期内,就复制到大多数。
2、如果随后的entry被提交。在上例中,如果s1持续成为term4的leader,那么entry2就会成为commit。

这是由于以下原因造成的:
更高任期为最新的投票规则,以及leader将其日志强加给follower。

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// updateCommitIndex find new commit id, must be called when hold lock
func (rf *Raft) updateCommitIndex() {
match := make([]int, len(rf.matchIndex))
copy(match, rf.matchIndex)
sort.Ints(match)

DPrintf("[%d-%s]: leader %d try to update commit index: %v @ term %d.\n",
rf.me, rf, rf.me, rf.matchIndex, rf.CurrentTerm)

target := match[len(rf.peers)/2]
if rf.commitIndex < target {
//fmt.Println("target:",target,match)
if rf.Logs[target].Term == rf.CurrentTerm {
//DPrintf("[%d-%s]: leader %d update commit index %d -> %d @ term %d command:%v\n",
// rf.me, rf, rf.me, rf.commitIndex, target, rf.CurrentTerm,rf.Logs[target].Command)

DPrintf("[%d-%s]: leader %d update commit index %d -> %d @ term %d\n",
rf.me, rf, rf.me, rf.commitIndex, target, rf.CurrentTerm)

rf.commitIndex = target
go func() { rf.commitCond.Broadcast() }()
} else {
DPrintf("[%d-%s]: leader %d update commit index %d failed (log term %d != current Term %d)\n",
rf.me, rf, rf.me, rf.commitIndex, rf.Logs[target].Term, rf.CurrentTerm)
}
}
}

参考

讲义
讲义新