User Contributed Notes
Referenci�k |
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20-Dec-2001 04:46 |
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Succesive assignement by reference and simple assignement :
$a =
'old';
$b = &$a;
$a = 'new';// now $b's value is new
$c =
$b;// simple copy
$a = 'new new'; // $b is 'new new', $c is still
'new'
And now with arrays :
$a[0] = 'old';
$b =
&$a;
$c = $b;//simple copy
$a[0] = 'new';// $b[0] is new, and
$c[0] is still 'old'
BUT :
$a = 'old';
$b[0] =
&$a;
$c = $b; // simple copy
$a = 'new';//now BOTH $b[0] and
$c[0] are 'new'
A simple copying of an array (or an object) does
NOT
break references for the values (or instance variables).
Use
var_dump() to see which of the values in an array
are
references.
Ivan
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[email protected]
11-Jan-2002 01:44 |
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You may be looking for how to iterate over an array by reference. For
example, you have an array of objects, and you want to call methods that
modify the objects. A normal foreach doesn't work, as the elements of the
array are copied. each() doesn't work either. You have to use a for
loop.
Example of what *COPIES* objects (not what you probably
want):
foreach($object_list as $object) {
$object->method();
}
Example of what does not copy objects
(probably what you want):
for ($c = 0; $c = count($object_list);
$c++) {
$object = &$object_list[$c];
$object->method();
}
You get the picture.
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[email protected]
17-Jan-2002 04:55 |
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Caveat, if your using an associative array with the key pointing to a
reference your going to want to do somehting like
this:
reset($MyArray);
while($Key = key($MyArray))
{
$_MyRef = & $MyArray[$Key];
...Do Smothing...
next($MyArray);
}
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cody at iensemble d0t c0m
07-Mar-2002 05:26 |
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Another way to iterate over an array of references:
foreach
(array_keys($myArray) as $key) {
doSomething($myArray[$key]);
}
This works on "normal"
and associative arrays.
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[email protected]_SPAM.nl
27-Mar-2002 10:37 |
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For the Java programmers among you, there's a big difference between Java
and PHP. In Java, all objects are passed by value (as in PHP). However,
try this in PHP:
class Object {
var $value;
function
Object() {
$this->value = 5;
}
function getValue()
{
return $this->value;
}
function setValue($value)
{
$this->value = $value;
}
}
function
updateObject($object) {
$object->setValue($object->getValue() +
5);
echo ', ', $object->getValue();
}
$object = new
Object();
echo
$object->getValue();
updateObject($object);
echo ', ',
$object->getValue();
It prints: '5, 10, 5'.
If you
implement this same code in Java however, it will print '5, 10,
10'!
Explanation: 'passing by value' in Java means that assigning a
different value to a variable inside a function will not change it outside
a function. (PHP equivalent: '$object = new Object'.) But calling methods
on the variable (when it's an object) can change the object (In PHP:
'$object->setValue(9)'). With PHP the latter isn't true,
unfortunately.
If you use objects in PHP, you have to use
references almost all the time, even if you don't want to assign a new
value to an outside variable, but just want to call methods (that might
update the object).
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[email protected]
02-Apr-2002 07:34 |
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If you want to check whether there is a reference between two variables you
can use this function.
function check_reference(&$var1,
&$var2)
{
// Save old values
$var1_old = $var1;
$var2_old = $var2;
// ($var1 . ".") must not be
$var2
if($var1 . "." == $var2)
$var1 .=
".";
// Change $var1
$var1 .=
".";
// Did $var2 also change?
if($var2 ==
$var1)
$reference = true;
else
$reference =
false;
// Old values
$var1 = $var1_old;
$var2 =
$var2_old;
return $reference;
}
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[email protected]
08-Apr-2002 09:34 |
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A have found interesting strange in references to classes.
class
Child{
function Child(&$parent)
{
$this->parent=&$parent;
$this->parent->var=10;
}
function doit()
{
$this->parent->var=20;
}
}
class A{
function A()
{
$this->child=new
Child($this);
}
function doit()
{
$this->child->doit();
echo $this->var;
}
}
$a = new A;
$a->doit();
-----------------
In
result we have 10, not 20.
And, if we call $a->doit(); from
constructor �():
function A()
{
$this->child=new
Child($this);
$this->doit();
}
we resulting 20.
The
way to takearound this is make empty constructor, move all from it to
function constr(), and run some like:
$a=new A;
$a->me=&$a;
$a->constr();
$a->doit();
in
constr() we have $this->child=new Child($this->me); [not
Child($this)].
and, as result we have 20 as would be
expected.
That's this? Am I missunderstand something or it is PHP
bug?
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23-Apr-2002 12:07 |
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When you do
$a = new A();
PHP creates an object for you,
and returns a COPY of that object
So (if I get it right... ) the child
is actually altering the original object and not the copy that you stored
in $a.
However if you do
$a =& new A();
PHP
still creates an object, but returns a REFERENCE to that object.
The
child is still altering the original - but you are happy, because you have
got a reference the original!
so if you change the line
$a = new A();
to
$a =& new A();
you'll get
the result you expected.
(BTW wouldn't it be more correct to say
that when you do:
$a = new A()
PHP returns not a copy
but a reference to a copy?)
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vic@NO_SPAM.altoona.net
16-Jul-2002 03:20 |
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Someone compared Java to PHP's pass by value. Java doesn't pass copies of
objects to methods "by value" as PHP does. It passes a copy of
the object's reference by value. It's still the same object. It's
similar to passing a pointer to a structure in C.
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[email protected]
16-Jul-2002 04:46 |
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Note that creating a reference to a non-existing array element causes that
element to be created with a null value. For
example:
var_dump($username); // => NULL
var_dump($session);
// => NULL
$username =&
$session['username'];
var_dump($username); // =>
NULL
var_dump($session);
// => array(1) {
["username"]=> &NULL }
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Referenci�k a konstruktorban