2019-07-27 20:00:12 +00:00
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<title>Lab: locks</title>
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<link rel="stylesheet" href="homework.css" type="text/css" />
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<body>
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<h1>Lab: locks</h1>
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<p>In this lab you will try to avoid lock contention for certain
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workloads.
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<h2>lock contention</h2>
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<p>The program user/kalloctest stresses xv6's memory allocator: three
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processes grow and shrink there address space, which will results in
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many calls to <tt>kalloc</tt> and <tt>kfree</tt>,
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respectively. <tt>kalloc</tt> and <tt>kfree</tt>
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obtain <tt>kmem.lock</tt>. To see if there is lock contention for
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<tt>kmem.lock</tt> replace the call to <tt>acquire</tt>
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in <tt>kalloc</tt> with the following code:
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<pre>
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while(!tryacquire(&kmem.lock)) {
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printf("!");
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}
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</pre>
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<p><tt>tryacquire</tt> tries to acquire <tt>kmem.lock</tt>: if the
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lock is taking it returns false (0); otherwise, it returns true (1)
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and with the lock acquired. Your first job is to
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implement <tt>tryacquire</tt> in kernel/spinlock.c.
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<p>A few hints:
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<ul>
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<li>look at <tt>acquire</tt>.
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<li>don't forget to restore interrupts when acquision fails
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<li>Add tryacquire's signature to defs.h.
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</ul>
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<p>Run usertests to see if you didn't break anything. Note that
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usertests never prints "!"; there is never contention
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for <tt>kmem.lock</tt>. The caller is always able to immediately
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acquire the lock and never has to wait because some other process
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has the lock.
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<p>Now run kalloctest. You should see quite a number of "!" on the
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console. kalloctest causes many processes to contend on
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the <tt>kmem.lock</tt>. This lock contention is a bit artificial,
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because qemu is simulating 3 processors, but it is likely on real
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hardware, there would be contention too.
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<h2>Removing lock contention</h2>
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<p>The root cause of lock contention in kalloctest is that there is a
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single free list, protected by a single lock. To remove lock
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contention, you will have to redesign the memory allocator to avoid
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a single lock and list. The basic idea is to maintain a free list
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per CPU, each list with its own lock. Allocations and frees on each
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CPU can run in parallel, because each CPU will operate on a
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different list.
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<p> The main challenge will be to deal with the case that one CPU runs
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out of memory, but another CPU has still free memory; in that case,
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the one CPU must "steal" part of the other CPU's free list.
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Stealing may introduce lock contention, but that may be acceptable
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because it may happen infrequently.
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<p>Your job is to implement per-CPU freelists and stealing when one
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CPU is out of memory. Run kalloctest() to see if your
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implementation has removed lock contention.
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<p>Some hints:
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<ul>
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2019-07-28 00:18:40 +00:00
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<li>You can use the constant <tt>NCPU</tt> in kernel/param.h
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<li>Let <tt>freerange</tt> give all free memory to the CPU
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running <tt>freerange</tt>.
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<li>The function <tt>cpuid</tt> returns the current core, but note
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that you can use it when interrupts are turned off and so you will
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need to turn on/off interrupts in your solution.
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2019-07-27 20:00:12 +00:00
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</ul>
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<p>Run usertests to see if you don't break anything.
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2019-07-28 01:02:37 +00:00
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2019-07-30 14:01:22 +00:00
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<h2>More scalabale bcache lookup</h2>
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2019-07-28 01:02:37 +00:00
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2019-07-29 19:49:47 +00:00
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<p>Several processes reading different files repeatedly will
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bottleneck in the buffer cache, bcache, in bio.c. Replace the
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acquire in <tt>bget</tt> with
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2019-07-28 01:02:37 +00:00
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2019-07-29 19:49:47 +00:00
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<pre>
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while(!tryacquire(&bcache.lock)) {
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printf("!");
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}
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</pre>
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and run test0 from bcachetest and you will see "!"s.
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<p>Modify <tt>bget</tt> so that a lookup for a buffer that is in the
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2019-07-30 17:07:17 +00:00
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bcache doesn't need to acquire <tt>bcache.lock</tt>. This is more
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2019-07-29 19:49:47 +00:00
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tricky than the kalloc assignment, because bcache buffers are truly
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shared among processes. You must maintain the invariant that a
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buffer is only once in memory.
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<p> There are several races that <tt>bcache.lock</tt> protects
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against, including:
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<ul>
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<li>A <tt>brelse</tt> may set <tt>b->ref</tt> to 0,
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while concurrent <tt>bget</tt> is incrementing it.
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<li>Two <tt>bget</tt> may see <tt>b->ref = 0</tt> and one may re-use
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the buffer, while the other may replaces it with another block.
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<li>A concurrent <tt>brelse</tt> modifies the list
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that <tt>bget</tt> traverses.
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</ul>
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<p>A challenge is testing whether you code is still correct. One way
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to do is to artificially delay certain operations
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2019-07-30 14:01:22 +00:00
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using <tt>sleepticks</tt>. <tt>test1</tt> trashes the buffer cache
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and exercises more code paths.
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2019-07-29 19:49:47 +00:00
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<p>Here are some hints:
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<ul>
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2019-07-30 17:07:17 +00:00
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<li>Read the description of buffer cache in the xv6 book (Section 7.2).
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2019-07-30 14:01:22 +00:00
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<li>Use a simple design: i.e., don't design a lock-free implementation.
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2019-07-30 17:07:17 +00:00
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<li>Use a simple hash table with locks per bucket.
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<li>Searching in hash table for a buffer and allocating an entry
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for that buffer when the buffer is not found must be atomic.
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<li>It is fine to acquire <tt>bcache.lock</tt> in <tt>brelse</tt>
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to update the LRU/MRU list.
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2019-07-29 19:49:47 +00:00
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</ul>
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2019-07-30 14:01:22 +00:00
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<p>Check that your implementation has less contention
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on <tt>test0</tt>
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<p>Make sure your implementation passes bcachetest and usertests.
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2019-07-30 17:07:17 +00:00
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<p>Optional:
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<ul>
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<li>make the buffer cache more scalable (e.g., avoid taking
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out <tt>bcache.lock</tt> on <tt>brelse</tt>).
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<li>make lookup lock-free (Hint: use gcc's <tt>__sync_*</tt>
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functions.) How do you convince yourself that your implementation is correct?
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</ul>
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2019-07-27 20:00:12 +00:00
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</body>
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</html>
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