Any difference between these two?

[Chung Leong]

Is there anything funcationally difference between the following
snippets:

// first
$lines[] = array();
$line =& $lines[count($lines) - 1];

// second
$line = array();
$lines[] =& $line;

They should do the thing. A voice in my head though is saying there's a
subtle difference.

[ZeldorBlat]

Chung Leong wrote:

Is there anything funcationally difference between the following
snippets:

// first
$lines[] = array();
$line =& $lines[count($lines) - 1];

// second
$line = array();
$lines[] =& $line;

They should do the thing. A voice in my head though is saying there's a
subtle difference.

They look the same to me -- at the end of the day both $line and
$lines[0] point to the same, empty array. Of course #2 is more
readable and straightforward.

[julian_m]

Chung Leong wrote:

// second
$line = array();
$lines[] =& $line;

Can anyone tell me what is the 2nd line? I mean, the first one declares
an array named "line", but the 2nd one?
I'm comfused by & and [] there, and as you can imagine, i'm a kind of
beginner right now...

regards - julian

[NC]

Chung Leong wrote:

Is there anything funcationally difference between the following
snippets:

// first
$lines[] = array();
$line =& $lines[count($lines) - 1];

// second
$line = array();
$lines[] =& $line;

Yes. The first example assumes that the array is numeric and there are
no gaps in numbering. If there are gaps, $lines[count($lines) - 1]
will not return the last element in the array. Consider this snippet:

$lines = array('Zero', 'One', 'Two', 'Three', 'Four');
unset($lines[1]);
print_r($lines); /* Outputs:
Array
(
[0] => Zero
[2] => Two
[3] => Three
[4] => Four
) */
echo $lines[count($lines) - 1], "\r\n"; // outputs "Three"...

The second example makes no such assumption and simply adds another
element to $lines...

Cheers,
NC

[JDS]

On Tue, 06 Dec 2005 18:48:03 -0800, Chung Leong wrote:

funcationally

Is that like, as pertaining to a really fun vacation?

--
JDS | je*****@example.invalid
| http://www.newtnotes.com
DJMBS | http://newtnotes.com/doctor-jeff-master-brainsurgeon/

[Chung Leong]

ZeldorBlat wrote:

Chung Leong wrote:

Is there anything funcationally difference between the following
snippets:

// first
$lines[] = array();
$line =& $lines[count($lines) - 1];

// second
$line = array();
$lines[] =& $line;

They should do the thing. A voice in my head though is saying there's a
subtle difference.

They look the same to me -- at the end of the day both $line and
$lines[0] point to the same, empty array. Of course #2 is more
readable and straightforward.

I think you're right. My concern was that in the first block, an array
was added to $lines, while in the second, a reference was added.
Momentarily I forgot that references don't exist as discrete objects.

Thanks.

[ZeldorBlat]

Chung Leong wrote:

ZeldorBlat wrote:

Chung Leong wrote:

Is there anything funcationally difference between the following
snippets:

// first
$lines[] = array();
$line =& $lines[count($lines) - 1];

// second
$line = array();
$lines[] =& $line;

They should do the thing. A voice in my head though is saying there's a
subtle difference.

They look the same to me -- at the end of the day both $line and
$lines[0] point to the same, empty array. Of course #2 is more
readable and straightforward.

I think you're right. My concern was that in the first block, an array
was added to $lines, while in the second, a reference was added.
Momentarily I forgot that references don't exist as discrete objects.

Thanks.

Yeah -- I like PHP's reference model a lot, mainly for it's simplicity:
every single "variable" is just a "label" for some value in memory.
Creating a "reference" is really just a way of creating another label
for that same value in that same spot in memory. Of course there are
some rules about what is always assigned by reference and what is
always copied.

[Chung Leong]

ZeldorBlat wrote:

Yeah -- I like PHP's reference model a lot, mainly for it's simplicity:
every single "variable" is just a "label" for some value in memory.
Creating a "reference" is really just a way of creating another label
for that same value in that same spot in memory. Of course there are
some rules about what is always assigned by reference and what is
always copied.

Well, simple logic could leads to scenarios that aren't that easily
understandable. My original question arose out of this function I'm
writing, which returns a large array of arrays. Knowing that PHP's
copy-on-write mechanism creates copies when "references" are returned,
I didn't want to return a whole array-ful of them. The line $lines[] =&
$line did look like that's what being created.

[Chung Leong]

ZeldorBlat wrote:

Yeah -- I like PHP's reference model a lot, mainly for it's simplicity:
every single "variable" is just a "label" for some value in memory.
Creating a "reference" is really just a way of creating another label
for that same value in that same spot in memory. Of course there are
some rules about what is always assigned by reference and what is
always copied.

Well, simple logic could leads to scenarios that aren't that easily
understandable. My original question arose out of this function I'm
writing, which returns a large array of arrays. Knowing that PHP's
copy-on-write mechanism creates copies when "references" are returned,
I didn't want to return a whole array-ful of them. The line $lines[] =&
$line does give the impression that that's what being created.

[Colin Fine]

ZeldorBlat wrote:

Yeah -- I like PHP's reference model a lot, mainly for it's simplicity:
every single "variable" is just a "label" for some value in memory.
Creating a "reference" is really just a way of creating another label
for that same value in that same spot in memory. Of course there are
some rules about what is always assigned by reference and what is
always copied.

And they're ENORMOUSLY different between PHP 4 and PHP 5.

Colin

[Colin Fine]

julian_m wrote:

Chung Leong wrote:

// second
$line = array();
$lines[] =& $line;

Can anyone tell me what is the 2nd line? I mean, the first one declares
an array named "line", but the 2nd one?
I'm comfused by & and [] there, and as you can imagine, i'm a kind of
beginner right now...

regards - julian

The second line adds a *reference* to the variable $line to the end of
array $lines; or to put it another way, it makes $lines[0] another name
for the same thing as $line. This means that even if the code later said

$line = 22;

(perfectly valid, though odd - it simply forgets about the array)
$lines[0] will have value 22.

If the '&' weren't there, then after
$line = 22,

$lines[0] will still be an array.

Colin