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diff --git a/Documentation/RCU/Design/Requirements/Requirements.html b/Documentation/RCU/Design/Requirements/Requirements.html index c67a96a2a389..acdad96f78e9 100644 --- a/Documentation/RCU/Design/Requirements/Requirements.html +++ b/Documentation/RCU/Design/Requirements/Requirements.html @@ -1,5 +1,3 @@ -<!-- DO NOT HAND EDIT. --> -<!-- Instead, edit Requirements.htmlx and run 'sh htmlqqz.sh Requirements' --> <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"> <html> @@ -65,8 +63,8 @@ All that aside, here are the categories of currently known RCU requirements: <p> This is followed by a <a href="#Summary">summary</a>, -which is in turn followed by the inevitable -<a href="#Answers to Quick Quizzes">answers to the quick quizzes</a>. +however, the answers to each quick quiz immediately follows the quiz. +Select the big white space with your mouse to see the answer. <h2><a name="Fundamental Requirements">Fundamental Requirements</a></h2> @@ -153,13 +151,27 @@ Therefore, the outcome: </blockquote> cannot happen. -<p><a name="Quick Quiz 1"><b>Quick Quiz 1</b>:</a> -Wait a minute! -You said that updaters can make useful forward progress concurrently -with readers, but pre-existing readers will block -<tt>synchronize_rcu()</tt>!!! -Just who are you trying to fool??? -<br><a href="#qq1answer">Answer</a> +<table> +<tr><th> </th></tr> +<tr><th align="left">Quick Quiz:</th></tr> +<tr><td> + Wait a minute! + You said that updaters can make useful forward progress concurrently + with readers, but pre-existing readers will block + <tt>synchronize_rcu()</tt>!!! + Just who are you trying to fool??? +</td></tr> +<tr><th align="left">Answer:</th></tr> +<tr><td bgcolor="#ffffff"><font color="ffffff"> + First, if updaters do not wish to be blocked by readers, they can use + <tt>call_rcu()</tt> or <tt>kfree_rcu()</tt>, which will + be discussed later. + Second, even when using <tt>synchronize_rcu()</tt>, the other + update-side code does run concurrently with readers, whether + pre-existing or not. +</font></td></tr> +<tr><td> </td></tr> +</table> <p> This scenario resembles one of the first uses of RCU in @@ -210,9 +222,20 @@ to guarantee that <tt>do_something()</tt> never runs concurrently with <tt>recovery()</tt>, but with little or no synchronization overhead in <tt>do_something_dlm()</tt>. -<p><a name="Quick Quiz 2"><b>Quick Quiz 2</b>:</a> -Why is the <tt>synchronize_rcu()</tt> on line 28 needed? -<br><a href="#qq2answer">Answer</a> +<table> +<tr><th> </th></tr> +<tr><th align="left">Quick Quiz:</th></tr> +<tr><td> + Why is the <tt>synchronize_rcu()</tt> on line 28 needed? +</td></tr> +<tr><th align="left">Answer:</th></tr> +<tr><td bgcolor="#ffffff"><font color="ffffff"> + Without that extra grace period, memory reordering could result in + <tt>do_something_dlm()</tt> executing <tt>do_something()</tt> + concurrently with the last bits of <tt>recovery()</tt>. +</font></td></tr> +<tr><td> </td></tr> +</table> <p> In order to avoid fatal problems such as deadlocks, @@ -332,12 +355,27 @@ It also prevents any number of “interesting” compiler optimizations, for example, the use of <tt>gp</tt> as a scratch location immediately preceding the assignment. -<p><a name="Quick Quiz 3"><b>Quick Quiz 3</b>:</a> -But <tt>rcu_assign_pointer()</tt> does nothing to prevent the -two assignments to <tt>p->a</tt> and <tt>p->b</tt> -from being reordered. -Can't that also cause problems? -<br><a href="#qq3answer">Answer</a> +<table> +<tr><th> </th></tr> +<tr><th align="left">Quick Quiz:</th></tr> +<tr><td> + But <tt>rcu_assign_pointer()</tt> does nothing to prevent the + two assignments to <tt>p->a</tt> and <tt>p->b</tt> + from being reordered. + Can't that also cause problems? +</td></tr> +<tr><th align="left">Answer:</th></tr> +<tr><td bgcolor="#ffffff"><font color="ffffff"> + No, it cannot. + The readers cannot see either of these two fields until + the assignment to <tt>gp</tt>, by which time both fields are + fully initialized. + So reordering the assignments + to <tt>p->a</tt> and <tt>p->b</tt> cannot possibly + cause any problems. +</font></td></tr> +<tr><td> </td></tr> +</table> <p> It is tempting to assume that the reader need not do anything special @@ -494,11 +532,42 @@ The <tt>rcu_access_pointer()</tt> on line 6 is similar to code protected by the corresponding update-side lock. </ol> -<p><a name="Quick Quiz 4"><b>Quick Quiz 4</b>:</a> -Without the <tt>rcu_dereference()</tt> or the -<tt>rcu_access_pointer()</tt>, what destructive optimizations -might the compiler make use of? -<br><a href="#qq4answer">Answer</a> +<table> +<tr><th> </th></tr> +<tr><th align="left">Quick Quiz:</th></tr> +<tr><td> + Without the <tt>rcu_dereference()</tt> or the + <tt>rcu_access_pointer()</tt>, what destructive optimizations + might the compiler make use of? +</td></tr> +<tr><th align="left">Answer:</th></tr> +<tr><td bgcolor="#ffffff"><font color="ffffff"> + Let's start with what happens to <tt>do_something_gp()</tt> + if it fails to use <tt>rcu_dereference()</tt>. + It could reuse a value formerly fetched from this same pointer. + It could also fetch the pointer from <tt>gp</tt> in a byte-at-a-time + manner, resulting in <i>load tearing</i>, in turn resulting a bytewise + mash-up of two distince pointer values. + It might even use value-speculation optimizations, where it makes + a wrong guess, but by the time it gets around to checking the + value, an update has changed the pointer to match the wrong guess. + Too bad about any dereferences that returned pre-initialization garbage + in the meantime! + </font> + + <p><font color="ffffff"> + For <tt>remove_gp_synchronous()</tt>, as long as all modifications + to <tt>gp</tt> are carried out while holding <tt>gp_lock</tt>, + the above optimizations are harmless. + However, + with <tt>CONFIG_SPARSE_RCU_POINTER=y</tt>, + <tt>sparse</tt> will complain if you + define <tt>gp</tt> with <tt>__rcu</tt> and then + access it without using + either <tt>rcu_access_pointer()</tt> or <tt>rcu_dereference()</tt>. +</font></td></tr> +<tr><td> </td></tr> +</table> <p> In short, RCU's publish-subscribe guarantee is provided by the combination @@ -571,28 +640,156 @@ systems with more than one CPU: <tt>synchronize_rcu()</tt> migrates in the meantime. </ol> -<p><a name="Quick Quiz 5"><b>Quick Quiz 5</b>:</a> -Given that multiple CPUs can start RCU read-side critical sections -at any time without any ordering whatsoever, how can RCU possibly tell whether -or not a given RCU read-side critical section starts before a -given instance of <tt>synchronize_rcu()</tt>? -<br><a href="#qq5answer">Answer</a> - -<p><a name="Quick Quiz 6"><b>Quick Quiz 6</b>:</a> -The first and second guarantees require unbelievably strict ordering! -Are all these memory barriers <i> really</i> required? -<br><a href="#qq6answer">Answer</a> - -<p><a name="Quick Quiz 7"><b>Quick Quiz 7</b>:</a> -You claim that <tt>rcu_read_lock()</tt> and <tt>rcu_read_unlock()</tt> -generate absolutely no code in some kernel builds. -This means that the compiler might arbitrarily rearrange consecutive -RCU read-side critical sections. -Given such rearrangement, if a given RCU read-side critical section -is done, how can you be sure that all prior RCU read-side critical -sections are done? -Won't the compiler rearrangements make that impossible to determine? -<br><a href="#qq7answer">Answer</a> +<table> +<tr><th> </th></tr> +<tr><th align="left">Quick Quiz:</th></tr> +<tr><td> + Given that multiple CPUs can start RCU read-side critical sections + at any time without any ordering whatsoever, how can RCU possibly + tell whether or not a given RCU read-side critical section starts + before a given instance of <tt>synchronize_rcu()</tt>? +</td></tr> +<tr><th align="left">Answer:</th></tr> +<tr><td bgcolor="#ffffff"><font color="ffffff"> + If RCU cannot tell whether or not a given + RCU read-side critical section starts before a + given instance of <tt>synchronize_rcu()</tt>, + then it must assume that the RCU read-side critical section + started first. + In other words, a given instance of <tt>synchronize_rcu()</tt> + can avoid waiting on a given RCU read-side critical section only + if it can prove that <tt>synchronize_rcu()</tt> started first. +</font></td></tr> +<tr><td> </td></tr> +</table> + +<table> +<tr><th> </th></tr> +<tr><th align="left">Quick Quiz:</th></tr> +<tr><td> + The first and second guarantees require unbelievably strict ordering! + Are all these memory barriers <i> really</i> required? +</td></tr> +<tr><th align="left">Answer:</th></tr> +<tr><td bgcolor="#ffffff"><font color="ffffff"> + Yes, they really are required. + To see why the first guarantee is required, consider the following + sequence of events: + </font> + + <ol> + <li> <font color="ffffff"> + CPU 1: <tt>rcu_read_lock()</tt> + </font> + <li> <font color="ffffff"> + CPU 1: <tt>q = rcu_dereference(gp); + /* Very likely to return p. */</tt> + </font> + <li> <font color="ffffff"> + CPU 0: <tt>list_del_rcu(p);</tt> + </font> + <li> <font color="ffffff"> + CPU 0: <tt>synchronize_rcu()</tt> starts. + </font> + <li> <font color="ffffff"> + CPU 1: <tt>do_something_with(q->a); + /* No smp_mb(), so might happen after kfree(). */</tt> + </font> + <li> <font color="ffffff"> + CPU 1: <tt>rcu_read_unlock()</tt> + </font> + <li> <font color="ffffff"> + CPU 0: <tt>synchronize_rcu()</tt> returns. + </font> + <li> <font color="ffffff"> + CPU 0: <tt>kfree(p);</tt> + </font> + </ol> + + <p><font color="ffffff"> + Therefore, there absolutely must be a full memory barrier between the + end of the RCU read-side critical section and the end of the + grace period. + </font> + + <p><font color="ffffff"> + The sequence of events demonstrating the necessity of the second rule + is roughly similar: + </font> + + <ol> + <li> <font color="ffffff">CPU 0: <tt>list_del_rcu(p);</tt> + </font> + <li> <font color="ffffff">CPU 0: <tt>synchronize_rcu()</tt> starts. + </font> + <li> <font color="ffffff">CPU 1: <tt>rcu_read_lock()</tt> + </font> + <li> <font color="ffffff">CPU 1: <tt>q = rcu_dereference(gp); + /* Might return p if no memory barrier. */</tt> + </font> + <li> <font color="ffffff">CPU 0: <tt>synchronize_rcu()</tt> returns. + </font> + <li> <font color="ffffff">CPU 0: <tt>kfree(p);</tt> + </font> + <li> <font color="ffffff"> + CPU 1: <tt>do_something_with(q->a); /* Boom!!! */</tt> + </font> + <li> <font color="ffffff">CPU 1: <tt>rcu_read_unlock()</tt> + </font> + </ol> + + <p><font color="ffffff"> + And similarly, without a memory barrier between the beginning of the + grace period and the beginning of the RCU read-side critical section, + CPU 1 might end up accessing the freelist. + </font> + + <p><font color="ffffff"> + The “as if” rule of course applies, so that any + implementation that acts as if the appropriate memory barriers + were in place is a correct implementation. + That said, it is much easier to fool yourself into believing + that you have adhered to the as-if rule than it is to actually + adhere to it! +</font></td></tr> +<tr><td> </td></tr> +</table> + +<table> +<tr><th> </th></tr> +<tr><th align="left">Quick Quiz:</th></tr> +<tr><td> + You claim that <tt>rcu_read_lock()</tt> and <tt>rcu_read_unlock()</tt> + generate absolutely no code in some kernel builds. + This means that the compiler might arbitrarily rearrange consecutive + RCU read-side critical sections. + Given such rearrangement, if a given RCU read-side critical section + is done, how can you be sure that all prior RCU read-side critical + sections are done? + Won't the compiler rearrangements make that impossible to determine? +</td></tr> +<tr><th align="left">Answer:</th></tr> +<tr><td bgcolor="#ffffff"><font color="ffffff"> + In cases where <tt>rcu_read_lock()</tt> and <tt>rcu_read_unlock()</tt> + generate absolutely no code, RCU infers quiescent states only at + special locations, for example, within the scheduler. + Because calls to <tt>schedule()</tt> had better prevent calling-code + accesses to shared variables from being rearranged across the call to + <tt>schedule()</tt>, if RCU detects the end of a given RCU read-side + critical section, it will necessarily detect the end of all prior + RCU read-side critical sections, no matter how aggressively the + compiler scrambles the code. + </font> + + <p><font color="ffffff"> + Again, this all assumes that the compiler cannot scramble code across + calls to the scheduler, out of interrupt handlers, into the idle loop, + into user-mode code, and so on. + But if your kernel build allows that sort of scrambling, you have broken + far more than just RCU! +</font></td></tr> +<tr><td> </td></tr> +</table> <p> Note that these memory-barrier requirements do not replace the fundamental @@ -637,9 +834,19 @@ inconvenience can be avoided through use of the <tt>call_rcu()</tt> and <tt>kfree_rcu()</tt> API members described later in this document. -<p><a name="Quick Quiz 8"><b>Quick Quiz 8</b>:</a> -But how does the upgrade-to-write operation exclude other readers? -<br><a href="#qq8answer">Answer</a> +<table> +<tr><th> </th></tr> +<tr><th align="left">Quick Quiz:</th></tr> +<tr><td> + But how does the upgrade-to-write operation exclude other readers? +</td></tr> +<tr><th align="left">Answer:</th></tr> +<tr><td bgcolor="#ffffff"><font color="ffffff"> + It doesn't, just like normal RCU updates, which also do not exclude + RCU readers. +</font></td></tr> +<tr><td> </td></tr> +</table> <p> This guarantee allows lookup code to be shared between read-side @@ -725,9 +932,20 @@ to do significant reordering. This is by design: Any significant ordering constraints would slow down these fast-path APIs. -<p><a name="Quick Quiz 9"><b>Quick Quiz 9</b>:</a> -Can't the compiler also reorder this code? -<br><a href="#qq9answer">Answer</a> +<table> +<tr><th> </th></tr> +<tr><th align="left">Quick Quiz:</th></tr> +<tr><td> + Can't the compiler also reorder this code? +</td></tr> +<tr><th align="left">Answer:</th></tr> +<tr><td bgcolor="#ffffff"><font color="ffffff"> + No, the volatile casts in <tt>READ_ONCE()</tt> and + <tt>WRITE_ONCE()</tt> prevent the compiler from reordering in + this particular case. +</font></td></tr> +<tr><td> </td></tr> +</table> <h3><a name="Readers Do Not Exclude Updaters">Readers Do Not Exclude Updaters</a></h3> @@ -780,10 +998,25 @@ new readers can start immediately after <tt>synchronize_rcu()</tt> starts, and <tt>synchronize_rcu()</tt> is under no obligation to wait for these new readers. -<p><a name="Quick Quiz 10"><b>Quick Quiz 10</b>:</a> -Suppose that synchronize_rcu() did wait until all readers had completed. -Would the updater be able to rely on this? -<br><a href="#qq10answer">Answer</a> +<table> +<tr><th> </th></tr> +<tr><th align="left">Quick Quiz:</th></tr> +<tr><td> + Suppose that synchronize_rcu() did wait until all readers had completed. + Would the updater be able to rely on this? +</td></tr> +<tr><th align="left">Answer:</th></tr> +<tr><td bgcolor="#ffffff"><font color="ffffff"> + No. + Even if <tt>synchronize_rcu()</tt> were to wait until + all readers had completed, a new reader might start immediately after + <tt>synchronize_rcu()</tt> completed. + Therefore, the code following + <tt>synchronize_rcu()</tt> cannot rely on there being no readers + in any case. +</font></td></tr> +<tr><td> </td></tr> +</table> <h3><a name="Grace Periods Don't Partition Read-Side Critical Sections"> Grace Periods Don't Partition Read-Side Critical Sections</a></h3> @@ -980,11 +1213,24 @@ grace period. As a result, an RCU read-side critical section cannot partition a pair of RCU grace periods. -<p><a name="Quick Quiz 11"><b>Quick Quiz 11</b>:</a> -How long a sequence of grace periods, each separated by an RCU read-side -critical section, would be required to partition the RCU read-side -critical sections at the beginning and end of the chain? -<br><a href="#qq11answer">Answer</a> +<table> +<tr><th> </th></tr> +<tr><th align="left">Quick Quiz:</th></tr> +<tr><td> + How long a sequence of grace periods, each separated by an RCU + read-side critical section, would be required to partition the RCU + read-side critical sections at the beginning and end of the chain? +</td></tr> +<tr><th align="left">Answer:</th></tr> +<tr><td bgcolor="#ffffff"><font color="ffffff"> + In theory, an infinite number. + In practice, an unknown number that is sensitive to both implementation + details and timing considerations. + Therefore, even in practice, RCU users must abide by the + theoretical rather than the practical answer. +</font></td></tr> +<tr><td> </td></tr> +</table> <h3><a name="Disabling Preemption Does Not Block Grace Periods"> Disabling Preemption Does Not Block Grace Periods</a></h3> @@ -1153,9 +1399,43 @@ synchronization primitives be legal within RCU read-side critical sections, including spinlocks, sequence locks, atomic operations, reference counters, and memory barriers. -<p><a name="Quick Quiz 12"><b>Quick Quiz 12</b>:</a> -What about sleeping locks? -<br><a href="#qq12answer">Answer</a> +<table> +<tr><th> </th></tr> +<tr><th align="left">Quick Quiz:</th></tr> +<tr><td> + What about sleeping locks? +</td></tr> +<tr><th align="left">Answer:</th></tr> +<tr><td bgcolor="#ffffff"><font color="ffffff"> + These are forbidden within Linux-kernel RCU read-side critical + sections because it is not legal to place a quiescent state + (in this case, voluntary context switch) within an RCU read-side + critical section. + However, sleeping locks may be used within userspace RCU read-side + critical sections, and also within Linux-kernel sleepable RCU + <a href="#Sleepable RCU"><font color="ffffff">(SRCU)</font></a> + read-side critical sections. + In addition, the -rt patchset turns spinlocks into a + sleeping locks so that the corresponding critical sections + can be preempted, which also means that these sleeplockified + spinlocks (but not other sleeping locks!) may be acquire within + -rt-Linux-kernel RCU read-side critical sections. + </font> + + <p><font color="ffffff"> + Note that it <i>is</i> legal for a normal RCU read-side + critical section to conditionally acquire a sleeping locks + (as in <tt>mutex_trylock()</tt>), but only as long as it does + not loop indefinitely attempting to conditionally acquire that + sleeping locks. + The key point is that things like <tt>mutex_trylock()</tt> + either return with the mutex held, or return an error indication if + the mutex was not immediately available. + Either way, <tt>mutex_trylock()</tt> returns immediately without + sleeping. +</font></td></tr> +<tr><td> </td></tr> +</table> <p> It often comes as a surprise that many algorithms do not require a @@ -1378,12 +1658,27 @@ write an RCU callback function that takes too long. Long-running operations should be relegated to separate threads or (in the Linux kernel) workqueues. -<p><a name="Quick Quiz 13"><b>Quick Quiz 13</b>:</a> -Why does line 19 use <tt>rcu_access_pointer()</tt>? -After all, <tt>call_rcu()</tt> on line 25 stores into the -structure, which would interact badly with concurrent insertions. -Doesn't this mean that <tt>rcu_dereference()</tt> is required? -<br><a href="#qq13answer">Answer</a> +<table> +<tr><th> </th></tr> +<tr><th align="left">Quick Quiz:</th></tr> +<tr><td> + Why does line 19 use <tt>rcu_access_pointer()</tt>? + After all, <tt>call_rcu()</tt> on line 25 stores into the + structure, which would interact badly with concurrent insertions. + Doesn't this mean that <tt>rcu_dereference()</tt> is required? +</td></tr> +<tr><th align="left">Answer:</th></tr> +<tr><td bgcolor="#ffffff"><font color="ffffff"> + Presumably the <tt>->gp_lock</tt> acquired on line 18 excludes + any changes, including any insertions that <tt>rcu_dereference()</tt> + would protect against. + Therefore, any insertions will be delayed until after + <tt>->gp_lock</tt> + is released on line 25, which in turn means that + <tt>rcu_access_pointer()</tt> suffices. +</font></td></tr> +<tr><td> </td></tr> +</table> <p> However, all that <tt>remove_gp_cb()</tt> is doing is @@ -1430,14 +1725,31 @@ This was due to the fact that RCU was not heavily used within DYNIX/ptx, so the very few places that needed something like <tt>synchronize_rcu()</tt> simply open-coded it. -<p><a name="Quick Quiz 14"><b>Quick Quiz 14</b>:</a> -Earlier it was claimed that <tt>call_rcu()</tt> and -<tt>kfree_rcu()</tt> allowed updaters to avoid being blocked -by readers. -But how can that be correct, given that the invocation of the callback -and the freeing of the memory (respectively) must still wait for -a grace period to elapse? -<br><a href="#qq14answer">Answer</a> +<table> +<tr><th> </th></tr> +<tr><th align="left">Quick Quiz:</th></tr> +<tr><td> + Earlier it was claimed that <tt>call_rcu()</tt> and + <tt>kfree_rcu()</tt> allowed updaters to avoid being blocked + by readers. + But how can that be correct, given that the invocation of the callback + and the freeing of the memory (respectively) must still wait for + a grace period to elapse? +</td></tr> +<tr><th align="left">Answer:</th></tr> +<tr><td bgcolor="#ffffff"><font color="ffffff"> + We could define things this way, but keep in mind that this sort of + definition would say that updates in garbage-collected languages + cannot complete until the next time the garbage collector runs, + which does not seem at all reasonable. + The key point is that in most cases, an updater using either + <tt>call_rcu()</tt> or <tt>kfree_rcu()</tt> can proceed to the + next update as soon as it has invoked <tt>call_rcu()</tt> or + <tt>kfree_rcu()</tt>, without having to wait for a subsequent + grace period. +</font></td></tr> +<tr><td> </td></tr> +</table> <p> But what if the updater must wait for the completion of code to be @@ -1862,11 +2174,26 @@ kthreads to be spawned. Therefore, invoking <tt>synchronize_rcu()</tt> during scheduler initialization can result in deadlock. -<p><a name="Quick Quiz 15"><b>Quick Quiz 15</b>:</a> -So what happens with <tt>synchronize_rcu()</tt> during -scheduler initialization for <tt>CONFIG_PREEMPT=n</tt> -kernels? -<br><a href="#qq15answer">Answer</a> +<table> +<tr><th> </th></tr> +<tr><th align="left">Quick Quiz:</th></tr> +<tr><td> + So what happens with <tt>synchronize_rcu()</tt> during + scheduler initialization for <tt>CONFIG_PREEMPT=n</tt> + kernels? +</td></tr> +<tr><th align="left">Answer:</th></tr> +<tr><td bgcolor="#ffffff"><font color="ffffff"> + In <tt>CONFIG_PREEMPT=n</tt> kernel, <tt>synchronize_rcu()</tt> + maps directly to <tt>synchronize_sched()</tt>. + Therefore, <tt>synchronize_rcu()</tt> works normally throughout + boot in <tt>CONFIG_PREEMPT=n</tt> kernels. + However, your code must also work in <tt>CONFIG_PREEMPT=y</tt> kernels, + so it is still necessary to avoid invoking <tt>synchronize_rcu()</tt> + during scheduler initialization. +</font></td></tr> +<tr><td> </td></tr> +</table> <p> I learned of these boot-time requirements as a result of a series of @@ -2571,10 +2898,23 @@ If you needed to wait on multiple different flavors of SRCU (but why???), you would need to create a wrapper function resembling <tt>call_my_srcu()</tt> for each SRCU flavor. -<p><a name="Quick Quiz 16"><b>Quick Quiz 16</b>:</a> -But what if I need to wait for multiple RCU flavors, but I also need -the grace periods to be expedited? -<br><a href="#qq16answer">Answer</a> +<table> +<tr><th> </th></tr> +<tr><th align="left">Quick Quiz:</th></tr> +<tr><td> + But what if I need to wait for multiple RCU flavors, but I also need + the grace periods to be expedited? +</td></tr> +<tr><th align="left">Answer:</th></tr> +<tr><td bgcolor="#ffffff"><font color="ffffff"> + If you are using expedited grace periods, there should be less penalty + for waiting on them in succession. + But if that is nevertheless a problem, you can use workqueues + or multiple kthreads to wait on the various expedited grace + periods concurrently. +</font></td></tr> +<tr><td> </td></tr> +</table> <p> Again, it is usually better to adjust the RCU read-side critical sections @@ -2678,377 +3018,4 @@ and is provided under the terms of the Creative Commons Attribution-Share Alike 3.0 United States license. -<h3><a name="Answers to Quick Quizzes"> -Answers to Quick Quizzes</a></h3> - -<a name="qq1answer"></a> -<p><b>Quick Quiz 1</b>: -Wait a minute! -You said that updaters can make useful forward progress concurrently -with readers, but pre-existing readers will block -<tt>synchronize_rcu()</tt>!!! -Just who are you trying to fool??? - - -</p><p><b>Answer</b>: -First, if updaters do not wish to be blocked by readers, they can use -<tt>call_rcu()</tt> or <tt>kfree_rcu()</tt>, which will -be discussed later. -Second, even when using <tt>synchronize_rcu()</tt>, the other -update-side code does run concurrently with readers, whether pre-existing -or not. - - -</p><p><a href="#Quick%20Quiz%201"><b>Back to Quick Quiz 1</b>.</a> - -<a name="qq2answer"></a> -<p><b>Quick Quiz 2</b>: -Why is the <tt>synchronize_rcu()</tt> on line 28 needed? - - -</p><p><b>Answer</b>: -Without that extra grace period, memory reordering could result in -<tt>do_something_dlm()</tt> executing <tt>do_something()</tt> -concurrently with the last bits of <tt>recovery()</tt>. - - -</p><p><a href="#Quick%20Quiz%202"><b>Back to Quick Quiz 2</b>.</a> - -<a name="qq3answer"></a> -<p><b>Quick Quiz 3</b>: -But <tt>rcu_assign_pointer()</tt> does nothing to prevent the -two assignments to <tt>p->a</tt> and <tt>p->b</tt> -from being reordered. -Can't that also cause problems? - - -</p><p><b>Answer</b>: -No, it cannot. -The readers cannot see either of these two fields until -the assignment to <tt>gp</tt>, by which time both fields are -fully initialized. -So reordering the assignments -to <tt>p->a</tt> and <tt>p->b</tt> cannot possibly -cause any problems. - - -</p><p><a href="#Quick%20Quiz%203"><b>Back to Quick Quiz 3</b>.</a> - -<a name="qq4answer"></a> -<p><b>Quick Quiz 4</b>: -Without the <tt>rcu_dereference()</tt> or the -<tt>rcu_access_pointer()</tt>, what destructive optimizations -might the compiler make use of? - - -</p><p><b>Answer</b>: -Let's start with what happens to <tt>do_something_gp()</tt> -if it fails to use <tt>rcu_dereference()</tt>. -It could reuse a value formerly fetched from this same pointer. -It could also fetch the pointer from <tt>gp</tt> in a byte-at-a-time -manner, resulting in <i>load tearing</i>, in turn resulting a bytewise -mash-up of two distince pointer values. -It might even use value-speculation optimizations, where it makes a wrong -guess, but by the time it gets around to checking the value, an update -has changed the pointer to match the wrong guess. -Too bad about any dereferences that returned pre-initialization garbage -in the meantime! - -<p> -For <tt>remove_gp_synchronous()</tt>, as long as all modifications -to <tt>gp</tt> are carried out while holding <tt>gp_lock</tt>, -the above optimizations are harmless. -However, -with <tt>CONFIG_SPARSE_RCU_POINTER=y</tt>, -<tt>sparse</tt> will complain if you -define <tt>gp</tt> with <tt>__rcu</tt> and then -access it without using -either <tt>rcu_access_pointer()</tt> or <tt>rcu_dereference()</tt>. - - -</p><p><a href="#Quick%20Quiz%204"><b>Back to Quick Quiz 4</b>.</a> - -<a name="qq5answer"></a> -<p><b>Quick Quiz 5</b>: -Given that multiple CPUs can start RCU read-side critical sections -at any time without any ordering whatsoever, how can RCU possibly tell whether -or not a given RCU read-side critical section starts before a -given instance of <tt>synchronize_rcu()</tt>? - - -</p><p><b>Answer</b>: -If RCU cannot tell whether or not a given -RCU read-side critical section starts before a -given instance of <tt>synchronize_rcu()</tt>, -then it must assume that the RCU read-side critical section -started first. -In other words, a given instance of <tt>synchronize_rcu()</tt> -can avoid waiting on a given RCU read-side critical section only -if it can prove that <tt>synchronize_rcu()</tt> started first. - - -</p><p><a href="#Quick%20Quiz%205"><b>Back to Quick Quiz 5</b>.</a> - -<a name="qq6answer"></a> -<p><b>Quick Quiz 6</b>: -The first and second guarantees require unbelievably strict ordering! -Are all these memory barriers <i> really</i> required? - - -</p><p><b>Answer</b>: -Yes, they really are required. -To see why the first guarantee is required, consider the following -sequence of events: - -<ol> -<li> CPU 1: <tt>rcu_read_lock()</tt> -<li> CPU 1: <tt>q = rcu_dereference(gp); - /* Very likely to return p. */</tt> -<li> CPU 0: <tt>list_del_rcu(p);</tt> -<li> CPU 0: <tt>synchronize_rcu()</tt> starts. -<li> CPU 1: <tt>do_something_with(q->a); - /* No smp_mb(), so might happen after kfree(). */</tt> -<li> CPU 1: <tt>rcu_read_unlock()</tt> -<li> CPU 0: <tt>synchronize_rcu()</tt> returns. -<li> CPU 0: <tt>kfree(p);</tt> -</ol> - -<p> -Therefore, there absolutely must be a full memory barrier between the -end of the RCU read-side critical section and the end of the -grace period. - -<p> -The sequence of events demonstrating the necessity of the second rule -is roughly similar: - -<ol> -<li> CPU 0: <tt>list_del_rcu(p);</tt> -<li> CPU 0: <tt>synchronize_rcu()</tt> starts. -<li> CPU 1: <tt>rcu_read_lock()</tt> -<li> CPU 1: <tt>q = rcu_dereference(gp); - /* Might return p if no memory barrier. */</tt> -<li> CPU 0: <tt>synchronize_rcu()</tt> returns. -<li> CPU 0: <tt>kfree(p);</tt> -<li> CPU 1: <tt>do_something_with(q->a); /* Boom!!! */</tt> -<li> CPU 1: <tt>rcu_read_unlock()</tt> -</ol> - -<p> -And similarly, without a memory barrier between the beginning of the -grace period and the beginning of the RCU read-side critical section, -CPU 1 might end up accessing the freelist. - -<p> -The “as if” rule of course applies, so that any implementation -that acts as if the appropriate memory barriers were in place is a -correct implementation. -That said, it is much easier to fool yourself into believing that you have -adhered to the as-if rule than it is to actually adhere to it! - - -</p><p><a href="#Quick%20Quiz%206"><b>Back to Quick Quiz 6</b>.</a> - -<a name="qq7answer"></a> -<p><b>Quick Quiz 7</b>: -You claim that <tt>rcu_read_lock()</tt> and <tt>rcu_read_unlock()</tt> -generate absolutely no code in some kernel builds. -This means that the compiler might arbitrarily rearrange consecutive -RCU read-side critical sections. -Given such rearrangement, if a given RCU read-side critical section -is done, how can you be sure that all prior RCU read-side critical -sections are done? -Won't the compiler rearrangements make that impossible to determine? - - -</p><p><b>Answer</b>: -In cases where <tt>rcu_read_lock()</tt> and <tt>rcu_read_unlock()</tt> -generate absolutely no code, RCU infers quiescent states only at -special locations, for example, within the scheduler. -Because calls to <tt>schedule()</tt> had better prevent calling-code -accesses to shared variables from being rearranged across the call to -<tt>schedule()</tt>, if RCU detects the end of a given RCU read-side -critical section, it will necessarily detect the end of all prior -RCU read-side critical sections, no matter how aggressively the -compiler scrambles the code. - -<p> -Again, this all assumes that the compiler cannot scramble code across -calls to the scheduler, out of interrupt handlers, into the idle loop, -into user-mode code, and so on. -But if your kernel build allows that sort of scrambling, you have broken -far more than just RCU! - - -</p><p><a href="#Quick%20Quiz%207"><b>Back to Quick Quiz 7</b>.</a> - -<a name="qq8answer"></a> -<p><b>Quick Quiz 8</b>: -But how does the upgrade-to-write operation exclude other readers? - - -</p><p><b>Answer</b>: -It doesn't, just like normal RCU updates, which also do not exclude -RCU readers. - - -</p><p><a href="#Quick%20Quiz%208"><b>Back to Quick Quiz 8</b>.</a> - -<a name="qq9answer"></a> -<p><b>Quick Quiz 9</b>: -Can't the compiler also reorder this code? - - -</p><p><b>Answer</b>: -No, the volatile casts in <tt>READ_ONCE()</tt> and -<tt>WRITE_ONCE()</tt> prevent the compiler from reordering in -this particular case. - - -</p><p><a href="#Quick%20Quiz%209"><b>Back to Quick Quiz 9</b>.</a> - -<a name="qq10answer"></a> -<p><b>Quick Quiz 10</b>: -Suppose that synchronize_rcu() did wait until all readers had completed. -Would the updater be able to rely on this? - - -</p><p><b>Answer</b>: -No. -Even if <tt>synchronize_rcu()</tt> were to wait until -all readers had completed, a new reader might start immediately after -<tt>synchronize_rcu()</tt> completed. -Therefore, the code following -<tt>synchronize_rcu()</tt> cannot rely on there being no readers -in any case. - - -</p><p><a href="#Quick%20Quiz%2010"><b>Back to Quick Quiz 10</b>.</a> - -<a name="qq11answer"></a> -<p><b>Quick Quiz 11</b>: -How long a sequence of grace periods, each separated by an RCU read-side -critical section, would be required to partition the RCU read-side -critical sections at the beginning and end of the chain? - - -</p><p><b>Answer</b>: -In theory, an infinite number. -In practice, an unknown number that is sensitive to both implementation -details and timing considerations. -Therefore, even in practice, RCU users must abide by the theoretical rather -than the practical answer. - - -</p><p><a href="#Quick%20Quiz%2011"><b>Back to Quick Quiz 11</b>.</a> - -<a name="qq12answer"></a> -<p><b>Quick Quiz 12</b>: -What about sleeping locks? - - -</p><p><b>Answer</b>: -These are forbidden within Linux-kernel RCU read-side critical sections -because it is not legal to place a quiescent state (in this case, -voluntary context switch) within an RCU read-side critical section. -However, sleeping locks may be used within userspace RCU read-side critical -sections, and also within Linux-kernel sleepable RCU -<a href="#Sleepable RCU">(SRCU)</a> -read-side critical sections. -In addition, the -rt patchset turns spinlocks into a sleeping locks so -that the corresponding critical sections can be preempted, which -also means that these sleeplockified spinlocks (but not other sleeping locks!) -may be acquire within -rt-Linux-kernel RCU read-side critical sections. - -<p> -Note that it <i>is</i> legal for a normal RCU read-side critical section -to conditionally acquire a sleeping locks (as in <tt>mutex_trylock()</tt>), -but only as long as it does not loop indefinitely attempting to -conditionally acquire that sleeping locks. -The key point is that things like <tt>mutex_trylock()</tt> -either return with the mutex held, or return an error indication if -the mutex was not immediately available. -Either way, <tt>mutex_trylock()</tt> returns immediately without sleeping. - - -</p><p><a href="#Quick%20Quiz%2012"><b>Back to Quick Quiz 12</b>.</a> - -<a name="qq13answer"></a> -<p><b>Quick Quiz 13</b>: -Why does line 19 use <tt>rcu_access_pointer()</tt>? -After all, <tt>call_rcu()</tt> on line 25 stores into the -structure, which would interact badly with concurrent insertions. -Doesn't this mean that <tt>rcu_dereference()</tt> is required? - - -</p><p><b>Answer</b>: -Presumably the <tt>->gp_lock</tt> acquired on line 18 excludes -any changes, including any insertions that <tt>rcu_dereference()</tt> -would protect against. -Therefore, any insertions will be delayed until after <tt>->gp_lock</tt> -is released on line 25, which in turn means that -<tt>rcu_access_pointer()</tt> suffices. - - -</p><p><a href="#Quick%20Quiz%2013"><b>Back to Quick Quiz 13</b>.</a> - -<a name="qq14answer"></a> -<p><b>Quick Quiz 14</b>: -Earlier it was claimed that <tt>call_rcu()</tt> and -<tt>kfree_rcu()</tt> allowed updaters to avoid being blocked -by readers. -But how can that be correct, given that the invocation of the callback -and the freeing of the memory (respectively) must still wait for -a grace period to elapse? - - -</p><p><b>Answer</b>: -We could define things this way, but keep in mind that this sort of -definition would say that updates in garbage-collected languages -cannot complete until the next time the garbage collector runs, -which does not seem at all reasonable. -The key point is that in most cases, an updater using either -<tt>call_rcu()</tt> or <tt>kfree_rcu()</tt> can proceed to the -next update as soon as it has invoked <tt>call_rcu()</tt> or -<tt>kfree_rcu()</tt>, without having to wait for a subsequent -grace period. - - -</p><p><a href="#Quick%20Quiz%2014"><b>Back to Quick Quiz 14</b>.</a> - -<a name="qq15answer"></a> -<p><b>Quick Quiz 15</b>: -So what happens with <tt>synchronize_rcu()</tt> during -scheduler initialization for <tt>CONFIG_PREEMPT=n</tt> -kernels? - - -</p><p><b>Answer</b>: -In <tt>CONFIG_PREEMPT=n</tt> kernel, <tt>synchronize_rcu()</tt> -maps directly to <tt>synchronize_sched()</tt>. -Therefore, <tt>synchronize_rcu()</tt> works normally throughout -boot in <tt>CONFIG_PREEMPT=n</tt> kernels. -However, your code must also work in <tt>CONFIG_PREEMPT=y</tt> kernels, -so it is still necessary to avoid invoking <tt>synchronize_rcu()</tt> -during scheduler initialization. - - -</p><p><a href="#Quick%20Quiz%2015"><b>Back to Quick Quiz 15</b>.</a> - -<a name="qq16answer"></a> -<p><b>Quick Quiz 16</b>: -But what if I need to wait for multiple RCU flavors, but I also need -the grace periods to be expedited? - - -</p><p><b>Answer</b>: -If you are using expedited grace periods, there should be less penalty -for waiting on them in succession. -But if that is nevertheless a problem, you can use workqueues or multiple -kthreads to wait on the various expedited grace periods concurrently. - - -</p><p><a href="#Quick%20Quiz%2016"><b>Back to Quick Quiz 16</b>.</a> - - </body></html> |