It does the whole of $b. It has to, because when you change 'baz', a reference in
'bar' needs to change to point to the newly copied 'baz', so 'bar' is
written...and likewise 'foo' is written.
Ben.
On 19/01/11 5:45 PM, Larry Garfield wrote:
Hi folks. I have a question about the PHP runtime that I hope is appropriate
for this list. (If not, please thwap me gently; I bruise easily.)
I know PHP does copy-on-write. However, how "deeply" does it copy when
dealing with nested arrays?
This is probably easiest to explain with an example...
$a['foo']['bar']['baz'] = 1;
$a['foo']['bar']['bob'] = 1;
$a['foo']['bar']['narf'] = 1;
$a['foo']['poink']['narf'] = 1;
function test($b) {
// Assume each of the following lines in isolation...
// Does this copy just the one variable baz, or the full array?
$b['foo']['bar']['baz'] = 2;
// Does this copy $b, or just $b['foo']['poink']?
$b['foo']['poink']['stuff'] = 3;
return $b;
}
// I know this is wasteful; I'm trying to figure out just how wasteful.
$a = test($a);
test() in this case should take $b by reference, but I'm trying to determine
how much of a difference it is. (In practice my use case has a vastly larger
array, so any inefficiencies are multiplied.)
--Larry Garfield
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