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[nycphp-talk] Pass-by-value and lazy copy

Gary Mort bz-gmort at beezifies.com
Thu Nov 22 08:34:02 EST 2007


So, in the pass by reference talk, it was pointed out that if you do:

$a = 100;
$b = 100;
// At this point in time, $a and $b are still using the same memory to 
store their data
$b=50;
// Now b is assigned it's own memory area since it changed the data

How does this work with objects?  For example:

$a->foo = "Foo";
$a->foobar = "Foobar";

$b = $a;
// at this point in time, their using the same memory location for their 
variables

$b->foobar="New Foobar";
// Is the entire object for $b copied to a new memory location, or is 
just $b->foobar given it's own memory?



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