RE: [PHP] learning php - problem already

> -----Original Message-----
> From: John W. Holmes [mailto:holmes072000@charter.net]
> Sent: 29 July 2003 23:05
>
> Curt Zirzow wrote:
> > Ok... I'm getting the red pen out now :)
> [snip]
> > the each() function returns a one element array that the current
> > (internal) array pointer is pointing to and will return false if at
> > the end of the array.
>
> It actually returns a four element array (as per the manual).
>
> > the list() function (not really a function) takes an array on the
> > right side of the = operator and assigns each variable its value in
> > order returned from the array.
>
> Right
>
> > so with the example array(0 => 'one', 1 => 'two'), the
> initial internal
> > pointer is looking at the first item so when the while statement
> > evaluates the the statement the each() function returns:
> > 0 => 'one'
>
> The four element array will be
> 1 => 'one'
> value => 'one'
> 0 => 0
> key => 0

OK, some more red pen coming along....

The four-element array would actually be:

0=>0
1=>'one'
'key'=>0
'value'=>'one'

in that order. So...

>
> > This array gets returned to the list statement
> > list($k, $v)

the list takes the first 2 elements (0=>0, 1=>'one') and assigns their values to $k and $v respectively, giving $k==0, $v=='one' -- the remaining 2 elements are dropped because there's nothing to assign them to. If you cared to put 4 variables in the list() structure, thus:

list($k, $v, $key, $value) = each($a);

you would, in this case, now have $k==0, $v=='one', $key==0, $value=='one'.

Cheers!

Mike

---------------------------------------------------------------------
Mike Ford, Electronic Information Services Adviser,
Learning Support Services, Learning & Information Services,
JG125, James Graham Building, Leeds Metropolitan University,
Beckett Park, LEEDS, LS6 3QS, United Kingdom
Email: m.ford@lmu.ac.uk
Tel: +44 113 283 2600 extn 4730 Fax: +44 113 283 3211

From: "Ford, Mike [LSS]" <M.Ford@lmu.ac.uk>
> From: John W. Holmes [mailto:holmes072000@charter.net]
> > The four element array will be
> > 1 => 'one'
> > value => 'one'
> > 0 => 0
> > key => 0
>
> OK, some more red pen coming along....

Since we're whipping them out (red pens that is)

> The four-element array would actually be:
>
> 0=>0
> 1=>'one'
> 'key'=>0
> 'value'=>'one'

Nope. Try this example from the manual and you'll see that you get the order
I gave

$foo = array ("bob", "fred", "jussi", "jouni", "egon", "marliese");
$bar = each ($foo);
print_r($bar);

The order is pretty irrelevant, though.

> list($k, $v, $key, $value) = each($a);
>
> you would, in this case, now have $k==0, $v=='one', $key==0,
$value=='one'.

Nope.. try that and you'll get two notices about undefined offsets at 3 and
2. $key and $value will not have any value at all.

Not that this really matters, but since this is fun to argue about...

$a = array('a' => 'abc');

Since we know that each($a) will return what I gave above, you basically
have this:

list($k,$v,$key,$value) = array(1=>'abc', 'value'=>'abc', 0=>'a',
'key'=>'a');

So, how this works is that list starts with $value. $value is at position
number four, so since arrays start at zero, it's going to look in the array
that was passed for an element at [3]. Since there is no [3] in the passed
array, you get a NOTICE about undefined offset.

Next it moves on to $key and looks for element [2]. Again you get a warning
since there is no element [2] in the passed array.

Next is $v and list is looking for [1]. Since [1] does exist and has a value
of 'abc', now $v = 'abc'

Last is $k and [0] and you get $k = 'a'.

That's how it works. :)

That's why this code:

list($a, $b, $c, $d) = array(4=>'four', 3=>'three', 2=>'two', 1=>'one',
0=>'zero');
echo "$a, $b, $c, $d";

gives:

zero, one, two, three

as the result, even though you passed a five element array in reverse order.

---John Holmes...