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    <p><strong>Short Answer</strong></p> <p>The first time that <code>calloc</code> is called it is explicitly zeroing out the memory. While the next time that it is called it assumed that the memory returned from <code>mmap</code> is already zeroed out.</p> <p><strong>Details</strong></p> <p>Here's some of the things that I checked to come to this conclusion that you could try yourself if you wanted:</p> <ol> <li><p>Insert a <code>calloc</code> call before your first <code>ALLOC</code> call. You will see that after this the Time for Time A and Time B are the same.</p></li> <li><p>Use the <code>clock()</code> function to check how long each of the <code>ALLOC</code> calls take. In the case where they are both using <code>calloc</code> you will see that the first call takes much longer than the second one.</p></li> <li><p>Use <code>time</code> to time the execution time of the <code>calloc</code> version and the <code>USE_MMAP</code> version. When I did this I saw that the execution time for <code>USE_MMAP</code> was consistently slightly less.</p></li> <li><p>I ran with <code>strace -tt -T</code> which shows both the time of when the system call was made and how long it took. Here is part of the output:</p></li> </ol> <p>Strace output:</p> <pre><code>21:29:06.127536 mmap(NULL, 2000015360, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_ANONYMOUS, -1, 0) = 0x7fff806fd000 &lt;0.000014&gt; 21:29:07.778442 mmap(NULL, 2000015360, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_ANONYMOUS, -1, 0) = 0x7fff093a0000 &lt;0.000021&gt; 21:29:07.778563 times({tms_utime=63, tms_stime=102, tms_cutime=0, tms_cstime=0}) = 4324241005 &lt;0.000011&gt; </code></pre> <p>You can see that the first <code>mmap</code> call took <code>0.000014</code> seconds, but that about <code>1.5</code> seconds elapsed before the next system call. Then the second <code>mmap</code> call took <code>0.000021</code> seconds, and was followed by the <code>times</code> call a few hundred microsecond later.</p> <p>I also stepped through part of the application execution with <code>gdb</code> and saw that the first call to <code>calloc</code> resulted in numerous calls to <code>memset</code> while the second call to <code>calloc</code> did not make any calls to <code>memset</code>. You can see the source code for <code>calloc</code> <a href="http://www.eglibc.org/cgi-bin/viewvc.cgi/trunk/libc/malloc/malloc.c?revision=17701&amp;view=markup" rel="nofollow">here</a> (look for <code>__libc_calloc</code>) if you are interested. As for why <code>calloc</code> is doing the <code>memset</code> on the first call but not subsequent ones I don't know. But I feel fairly confident that this explains the behavior you have asked about.</p> <p>As for why the array that was zeroed <code>memset</code> has improved performance my guess is that it is because of values being loaded into the TLB rather than the cache since it is a very large array. Regardless the specific reason for the performance difference that you asked about is that the two <code>calloc</code> calls behave differently when they are executed. </p>
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    1. POCan some explain the performance behavior of the following memory allocating C program?
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    1. USGabriel Southern
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    1. POCan some explain the performance behavior of the following memory allocating C program?
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    1. COalso I originally thought that the behavior was related to the cache, but my guess now is the TLB or something else related to virtual memory and page allocation. The amount of memory allocated is so large that I don't think it will remain in the cache. Regardless I am confident that the performance difference is related to whether or not the memory is explicitly zeroed by `calloc`.
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    2. COSkimming the source I think the reason the second calloc may not need to zero any mem is because routine knows that mem freshly allocated to process by OS is already zeroed.
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    3. COThat's true, but it's also true of the first `calloc` call in this example. So really neither of them should have to call `memset`.
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