…ecution order test

In redis#12301, we observed that if the
`while(listLength(server.ready_keys) != 0)`
in handleClientsBlockedOnKeys is changed to
`if(listLength(server.ready_keys) != 0)`,
the order of command execution will change.

It is wrong to change that. It means that if a command
being unblocked causes another command to get unblocked
(like a BLMOVE would do), then the new unblocked command
will wait for later to get processed rather than right away.

It'll not have any real implication if we change that since
we do call handleClientsBlockedOnKeys in beforeSleep again,
and redis will still behave correctly, but we don't change that.

An example:
1. $rd1 blmove src{t} dst{t} left right 0
2. $rd2 blmove dst{t} src{t} right left 0
3. $rd3 set key1{t}, $rd3 lpush src{t}, $rd3 set key2{t} in a pipeline

The correct order would be:
1. set key1{t}
2. lpush src{t}
3. lmove src{t} dst{t} left right
4. lmove dst{t} src{t} right left
5. set key2{t}

The wrong order would be:
1. set key1{t}
2. lpush src{t}
3. lmove src{t} dst{t} left right
4. set key2{t}
5. lmove dst{t} src{t} right left

This PR adds corresponding test to cover it.

…ecution order test (#12324)

* Add command being unblocked cause another command to get unblocked execution order test

In #12301, we observed that if the
`while(listLength(server.ready_keys) != 0)`
in handleClientsBlockedOnKeys is changed to
`if(listLength(server.ready_keys) != 0)`,
the order of command execution will change.

It is wrong to change that. It means that if a command
being unblocked causes another command to get unblocked
(like a BLMOVE would do), then the new unblocked command
will wait for later to get processed rather than right away.

It'll not have any real implication if we change that since
we do call handleClientsBlockedOnKeys in beforeSleep again,
and redis will still behave correctly, but we don't change that.

An example:
1. $rd1 blmove src{t} dst{t} left right 0
2. $rd2 blmove dst{t} src{t} right left 0
3. $rd3 set key1{t}, $rd3 lpush src{t}, $rd3 set key2{t} in a pipeline

The correct order would be:
1. set key1{t}
2. lpush src{t}
3. lmove src{t} dst{t} left right
4. lmove dst{t} src{t} right left
5. set key2{t}

The wrong order would be:
1. set key1{t}
2. lpush src{t}
3. lmove src{t} dst{t} left right
4. set key2{t}
5. lmove dst{t} src{t} right left

This PR adds corresponding test to cover it.

* Add comment near while(listLength(server.ready_keys) != 0)

Apart from adding the missing coverage, this PR also adds `blockedBeforeSleep`
that gathers all block-related functions from `beforeSleep`

The order inside `blockedBeforeSleep` is different: now `handleClientsBlockedOnKeys`
(which may unblock clients) is called before `processUnblockedClients` (which handles
unblocked clients).
It makes sense to have this order.

There are no visible effects of the wrong ordering, except some cleanups of the now-unblocked
client would have  happen in the next `beforeSleep` (will now happen in the current one)

The reason we even got into it is because i triggers an assertion in logresreq.c (breaking
the assumption that `unblockClient` is called **before** actually flushing the reply to the socket):
`handleClientsBlockedOnKeys` is called, then it calls `moduleUnblockClientOnKey`, which calls
`moduleUnblockClient`, which adds the client to `moduleUnblockedClients` back to `beforeSleep`,
we call `handleClientsWithPendingWritesUsingThreads`, it writes the data of buf to the client, so
`client->bufpos` became 0
On the next `beforeSleep`, we call `moduleHandleBlockedClients`, which calls `unblockClient`,
which calls `reqresAppendResponse`, triggering the assert. (because the `bufpos` is 0) - see #12301 (comment)