Processing command from the master while the slave is in busy state is
not correct, however we cannot, also, just reply -BUSY to the
replication stream commands from the master. The correct solution is to
stop processing data from the master, but just accumulate the stream
into the buffers and resume the processing later.
Related to #5297.
To avoid copying buffers to create a large Redis Object which
exceeding PROTO_IOBUF_LEN 32KB, we just read the remaining data
we need, which may less than PROTO_IOBUF_LEN. But the remaining
len may be zero, if the bulklen+2 equals sdslen(c->querybuf),
in client pause context.
For example:
Time1:
python
>>> import os, socket
>>> server="127.0.0.1"
>>> port=6379
>>> data1="*3\r\n$3\r\nset\r\n$1\r\na\r\n$33000\r\n"
>>> data2="".join("x" for _ in range(33000)) + "\r\n"
>>> data3="\n\n"
>>> s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
>>> s.settimeout(10)
>>> s.connect((server, port))
>>> s.send(data1)
28
Time2:
redis-cli client pause 10000
Time3:
>>> s.send(data2)
33002
>>> s.send(data3)
2
>>> s.send(data3)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
socket.error: [Errno 104] Connection reset by peer
To fix that, we should check if remaining is greater than zero.
Note: this breaks backward compatibility with Redis 4, since now slaves
by default are exact copies of masters and do not try to evict keys
independently.
Function setProtocolError just records proctocol error
details in server log, set client as CLIENT_CLOSE_AFTER_REPLY.
It doesn't care about querybuf sdsrange, because we
will do it after procotol parsing.
This is an optimization for processing pipeline, we discussed a
problem in issue #5229: clients may be paused if we apply `CLIENT
PAUSE` command, and then querybuf may grow too large, the cost of
memmove in sdsrange after parsing a completed command will be
horrible. The optimization is that parsing all commands in queyrbuf
, after that we can just call sdsrange only once.
When the element new score is the same of prev/next node, the
lexicographical order kicks in, so we can safely update the node in
place only when the new score is strictly between the adjacent nodes
but never equal to one of them.
Technically speaking we could do extra checks to make sure that even if the
score is the same as one of the adjacent nodes, we can still update on
place, but this rarely happens, so probably not a good deal to make it
more complex.
Related to #5179.
