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| diff --git a/labs/lock.html b/labs/lock.html deleted file mode 100644 index 707d6c4..0000000 --- a/labs/lock.html +++ /dev/null @@ -1,148 +0,0 @@ -<html> -<head> -<title>Lab: locks</title> -<link rel="stylesheet" href="homework.css" type="text/css" /> -</head> -<body> - -<h1>Lab: locks</h1> - -<p>In this lab you will try to avoid lock contention for certain -workloads. - -<h2>lock contention</h2> - -<p>The program user/kalloctest stresses xv6's memory allocator: three -  processes grow and shrink there address space, which will results in -  many calls to <tt>kalloc</tt> and <tt>kfree</tt>, -  respectively.  <tt>kalloc</tt> and <tt>kfree</tt> -  obtain <tt>kmem.lock</tt>.  To see if there is lock contention for -  <tt>kmem.lock</tt> replace the call to <tt>acquire</tt> -  in <tt>kalloc</tt> with the following code: - -  <pre> -    while(!tryacquire(&kmem.lock)) { -      printf("!"); -    } -  </pre> - -<p><tt>tryacquire</tt> tries to acquire <tt>kmem.lock</tt>: if the -  lock is taking it returns false (0); otherwise, it returns true (1) -  and with the lock acquired.  Your first job is to -  implement <tt>tryacquire</tt> in kernel/spinlock.c. - -<p>A few hints: -  <ul> -    <li>look at <tt>acquire</tt>. -    <li>don't forget to restore interrupts when acquision fails -    <li>Add tryacquire's signature to defs.h. -  </ul> - -<p>Run usertests to see if you didn't break anything.  Note that -  usertests never prints "!"; there is never contention -  for <tt>kmem.lock</tt>.  The caller is always able to immediately -  acquire the lock and never has to wait because some other process -  has the lock. - -<p>Now run kalloctest.  You should see quite a number of "!" on the -  console.  kalloctest causes many processes to contend on -  the <tt>kmem.lock</tt>.  This lock contention is a bit artificial, -  because qemu is simulating 3 processors, but it is likely on real -  hardware, there would be contention too. -   -<h2>Removing lock contention</h2> - -<p>The root cause of lock contention in kalloctest is that there is a -  single free list, protected by a single lock.  To remove lock -  contention, you will have to redesign the memory allocator to avoid -  a single lock and list.  The basic idea is to maintain a free list -  per CPU, each list with its own lock. Allocations and frees on each -  CPU can run in parallel, because each CPU will operate on a -  different list. -   -<p> The main challenge will be to deal with the case that one CPU runs -  out of memory, but another CPU has still free memory; in that case, -  the one CPU must "steal" part of the other CPU's free list. -  Stealing may introduce lock contention, but that may be acceptable -  because it may happen infrequently. - -<p>Your job is to implement per-CPU freelists and stealing when one -  CPU is out of memory.  Run kalloctest() to see if your -  implementation has removed lock contention. - -<p>Some hints: -  <ul> -    <li>You can use the constant <tt>NCPU</tt> in kernel/param.h -    <li>Let <tt>freerange</tt> give all free memory to the CPU -      running <tt>freerange</tt>. -    <li>The function <tt>cpuid</tt> returns the current core, but note -    that you can use it when interrupts are turned off and so you will -    need to turn on/off interrupts in your solution. -  </ul> - -<p>Run usertests to see if you don't break anything. - -<h2>More scalabale bcache lookup</h2> - - -<p>Several processes reading different files repeatedly will -  bottleneck in the buffer cache, bcache, in bio.c.  Replace the -  acquire in <tt>bget</tt> with -   -  <pre> -    while(!tryacquire(&bcache.lock)) { -      printf("!"); -    } -  </pre> - -  and run test0 from bcachetest and you will see "!"s. - -<p>Modify <tt>bget</tt> so that a lookup for a buffer that is in the -  bcache doesn't need to acquire <tt>bcache.lock</tt>.  This is more -  tricky than the kalloc assignment, because bcache buffers are truly -  shared among processes. You must maintain the invariant that a -  buffer is only once in memory. - -<p> There are several races that <tt>bcache.lock</tt> protects -against, including: -  <ul> -    <li>A <tt>brelse</tt> may set <tt>b->ref</tt> to 0, -      while concurrent <tt>bget</tt> is incrementing it. -    <li>Two <tt>bget</tt> may see <tt>b->ref = 0</tt> and one may re-use -    the buffer, while the other may replaces it with another block. -    <li>A concurrent <tt>brelse</tt> modifies the list -      that <tt>bget</tt> traverses. -  </ul> - -<p>A challenge is testing whether you code is still correct.  One way -  to do is to artificially delay certain operations -  using <tt>sleepticks</tt>.  <tt>test1</tt> trashes the buffer cache -  and exercises more code paths. - -<p>Here are some hints: -  <ul> -    <li>Read the description of buffer cache in the xv6 book (Section 7.2). -    <li>Use a simple design: i.e., don't design a lock-free implementation. -    <li>Use a simple hash table with locks per bucket. -    <li>Searching in hash table for a buffer and allocating an entry -      for that buffer when the buffer is not found must be atomic. -    <li>It is fine to acquire <tt>bcache.lock</tt> in <tt>brelse</tt> -      to update the LRU/MRU list. -  </ul> - -<p>Check that your implementation has less contention -  on <tt>test0</tt> - -<p>Make sure your implementation passes bcachetest and usertests. - -<p>Optional: -  <ul> -  <li>make the buffer cache more scalable (e.g., avoid taking -  out <tt>bcache.lock</tt> on <tt>brelse</tt>). -  <li>make lookup lock-free (Hint: use gcc's <tt>__sync_*</tt> -    functions.) How do you convince yourself that your implementation is correct? -  </ul> -   -   -</body> -</html> | 
