130 lines
		
	
	
	
		
			4.5 KiB
		
	
	
	
		
			HTML
		
	
	
	
	
	
			
		
		
	
	
			130 lines
		
	
	
	
		
			4.5 KiB
		
	
	
	
		
			HTML
		
	
	
	
	
	
<html>
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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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</head>
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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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    <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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  </ul>
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<p>Run usertests to see if you don't break anything.
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<h2>Lock-free bcache lookup</h2>
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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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  <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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  bcache doesn't need to acquire <tt>bcache.lock</tt>.  This more
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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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  using <tt>sleepticks</tt>.
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<p>Here are some hints:
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  <ul>
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    <li> Use an atomic increment instruction for incrementing and
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      decrementing <tt>b->ref</tt> (see <tt>__sync_fetch_and_add() and
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	related primitives</tt>)
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    <li>Don't walk the <tt>bcache.head</tt> list to find a buffer
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  </ul>
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</body>
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</html>
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