Comments (4)
不好意思上班比较忙,晚上回去我详细看一下
from 6.824.
不好意思上周忘记了,是没问题的兄弟,你看下第525行,leader为每个server维护的nextindex是不一样的,发出的rpc请求中的log是不一样长的,对于你说的宕机server会从0开始。
from 6.824.
不好意思上周忘记了,是没问题的兄弟,你看下第525行,leader为每个server维护的nextindex是不一样的,发出的rpc请求中的log是不一样长的,对于你说的宕机server会从0开始。
但是你在337行,每个新leader出现后,会把所有nextIndex设置为最新idx,这样在下一个appendEntry就会出现贴主说的问题。
按照原论文的实现,appendEntry应该是每次单步的情形从后向前试错的,只有在找到第一个idx和term都相同的节点才会进行append操作,所以这种一开始log.size()==0的情况我觉得好像得特化处理。
from 6.824.
是这样的,后续论文里我记得给出了优化的方案,是快速回滚的一种方法,MIT的资料里应该也有这部分。所以我是按照那个逻辑来,具体的记不太清楚了,周末我再看一下
from 6.824.
Related Issues (8)
- 在Lab1的master.cpp中的多线程定时器效率太低 HOT 4
- 请问这个对应哪一年那一季度的课程呢
- lab1中map执行失败的任务无法重新进行分配是啥原因?
- raft.cpp 函数中的代码:https://github.com/tjumcw/6.824/blob/991d662ceaaa7308142f6103c4dd024ce49f534c/src/Raft/raft.cpp#L588是否写错了 HOT 3
- Feature request : Improving the implementation of timers in Lab1 HOT 5
- 关于shardkv中的问题 HOT 2
- 交流
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