Req #51280 [Com]: Calculate expression before using it as argument
Edit report at http://bugs.php.net/bug.php?id=51280&edit=1
ID: 51280
Comment by: olamedia at gmail dot com
Reported by: olamedia at gmail dot com
Summary: Calculate expression before using it as argument
Status: Bogus
Type: Feature/Change Request
Package: *General Issues
PHP Version: 5.3.2
New Comment:
well, summarizing, I think that
1. any expression MUST be evaluated before it can be used (as parameter
of
function or anything).
2. it's buggy that __set() is not working in this special case.
Previous Comments:
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[2010-03-22 10:30:32] olamedia at gmail dot com
Precalculation can be useful for ORM, for example, giving ability to use
this
construction:
$users->where($group->id = 2)
Also, this can be used for named parameters emulation:
class parameter{
var $name;
var $value;
function __construct($name, $value){
$this->name = $name;
$this->value = $value;
}
}
class parameters{
var $data = array();
function __set($name, $value){
$this->data[$name] = new parameter($name, $value);
}
function __get($name){
return $this->data[$name];
}
}
function a(){
var_dump(func_get_args());
}
$arg = new parameters();
a($arg->id = 3, $arg->x = 'some', $arg->y = 5);
------------------------------------------------------------------------
[2010-03-22 10:18:40] johannes@php.net
The order of the evaluation of function parameters is "undefined". (It
actually changed with 5.2 or so due to performance reasons)
------------------------------------------------------------------------
[2010-03-22 09:51:47] olamedia at gmail dot com
Sorry, my bad. This was an incomplete example. In first message there
was a good
example, when $a->x is an object, and in both cases: a($a->x) and
a($a->x = 3) - I
'm expecting object $a->x instead of assigned value because of __set()
function,
which converts value to an object.
------------------------------------------------------------------------
[2010-03-22 09:43:51] olamedia at gmail dot com
> the value of "$a = 3" is 3
Here I'm talking about function arguments, not a simple assignments.
function b($a){
// here expected resulting object $a, not just int(3)
var_dump($a);
}
class a{}
$a = new a();
b($a->x = 3); // outputs int(3)
$a->x = 3
b($a); // outputs object(a)#2 (1) { ["x"]=> int(3) }
------------------------------------------------------------------------
[2010-03-21 21:07:08] hholzgra@php.net
Expected behavior, see:
http://php.net/manual/en/language.operators.assignment.php
[...]
The value of an assignment expression is the value assigned.
That is, the value of "$a = 3" is 3.
[...]
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