It is said that PHP uses copy-on-write processes. Then I wander if I run these codes:
$first = 5;
$second = $first;
$first = 5;
Then does it allocate new memory space for $first? Many thanks
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2 Answers
run this script twice. first time:
echo "<pre>";
$first = 5;
echo memory_get_usage() . "\n";
$second = $first;
echo memory_get_usage() . "\n";
$first = 5;
echo memory_get_usage() . "\n";
result:
333224
333280
333312
second time - just comment one line
echo "<pre>";
$first = 5;
echo memory_get_usage() . "\n";
//$second = $first;
echo memory_get_usage() . "\n";
$first = 5;
echo memory_get_usage() . "\n";
result:
333112
333112
333112
answer: yes, it allocates new memory
Comments
The semantic is "copy-on-write" NOT "copy-on-write-only-if-the-value-has-changed".
Also some space for $second is created as soon you declare it. The minimal space required for that variable to exist. At this time the space is NOT allocated for the value that the $first has. So copy on write is at work here; there is no write so no extra space allocated for the value allocated to $first, which is now being assigned to $second.
Then further space is allocated to $second the moment you assign something else to $first. At that time the space is created to hold the copy of the original value in $first
<?php
echo "<pre>";
// 10 sets == 100 chars
$first = "1234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890";
echo memory_get_usage() . "\n";
$second = $first;
echo memory_get_usage() . "\n";
$first = "1234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890";
echo memory_get_usage() . "\n";
$first = "1234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890";
echo memory_get_usage() . "\n";
Output
241496
241584
241752
241752
