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Re: Last element in an array? [message #169459 is a reply to message #169457] Mon, 13 September 2010 21:53 Go to previous messageGo to previous message
Marious Barrier is currently offline  Marious Barrier
Messages: 25
Registered: September 2010
Karma:
Junior Member
On 09/13/2010 05:34 PM, matt wrote:
> On Sep 13, 2:56 pm, Marious Barrier<marious.barr...@gmail.com> wrote:
>> On 09/13/2010 02:14 PM, matt wrote:
>>
>>
>>
>>> On Sep 11, 3:32 pm, Thomas Mlynarczyk<tho...@mlynarczyk-webdesign.de>
>>> wrote:
>>>> MikeB schrieb:
>>
>>>> > I know of the end() function to find the last element in an array [...]
>>>> > What I can't seem to find is a function that returns the key value of
>>>> > the last element.
>>
>>>> function lastKey( $array )
>>>> {
>>>> end( $array );
>>>> return key( $array );
>>
>>>> }
>>
>>>> Greetings,
>>>> Thomas
>>
>>> I'm curious. Will that affect the internal array pointer in all
>>> scopes? ie,
>>
>>> $array = array(0, 1, 2);
>>> next($array); // internal pointer -> 1
>>> lastKey($array);
>>> echo current($array); // internal pointer -> 2 ??
>>
>> Try it to be sure.
>> It should not... when not passing an array by reference, you are working
>> with a “copy” of the array and its internal pointer.
>
> Right...I believe that however, if you do this:
>
> $a = array(str_repeat("A", 10000));
> echo memory_get_usage() . "\n";
> $b = $a; // a full copy of a has NOT been made yet
> echo memory_get_usage() . "\n";
> $b[] = "B"; // now it has, I believe
> echo memory_get_usage() . "\n";
>
> Because PHP is doing some sort of smart memory management here that I
> don't fully understand. My results for this actually don't seem to
> line up with my understanding of how and when PHP copies arrays. I
> got the following:
>
> 328436
> 328504
> 328752
>
> I would have expected the third number to be almost double the first
> based on my understanding. If PHP were making a copy right at the
> assignment operator, then I'd expect to see the 2nd and 3rd be double
> the first.
>
> Yet, the question remains, do the array pointer functions trigger PHP
> to make a full copy of the array, or at least maintain separate
> pointers to the same memory space?
>
> $a = array(0, 1, 2);
> $b = $a;
> next($b);
> printf("%d, %d\n", current($a), current($b));
>
> yields: "0,1"
>
> So, the answer seems that yes, that does seem to be enough to trigger
> PHP to make a full copy, or that $a and $b maintain separate
> pointers. Perhaps someone with a deeper knowledge of the internals
> could help convince me that I am interpreting these results correctly.

Its simple, PHP won’t make a real copy in memory when there’s no need.
When you do not pass a variable by reference, the working data will be
the same, like if it was a reference, until you change it.
Try doing it.
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