Hello,
I've upgraded from 5.1.1 to 5.1.2 and discovered the following
unexpected difference. Best shown by example.
<?
function getArray(&$arr)
{
$arr[] = 12;
}
getArray($p = array());
print_r($p);
?>
In php 5.1.1, the $p is passed correctly as reference and so returns
with the value 12 in its first element. I regard this as correct behaviour.
In php 5.1.2, the $p is passed by value. So on return, $p is still empty!
This can be fixed as follows:
$p = array();
getArray($p);
The following example works, as expected, in both 5.1.1 and 5.1.2.
<?
class A
{
public $i = 9;
}
function doAmend(A &$a) // & not required because class instance passed
by reference by default
{
$a->i += 30;
}
doAmend($b = new A);
print_r($b); // this prints the value of member i as 39 as
expected.
?
Hello,
I've upgraded from 5.1.1 to 5.1.2 and discovered the following
unexpected difference. Best shown by example.<?
function getArray(&$arr)
{
$arr[] = 12;
}getArray($p = array()); print_r($p);?>
In php 5.1.1, the $p is passed correctly as reference and so returns
with the value 12 in its first element. I regard this as correct
behaviour. In php 5.1.2, the $p is passed by value. So on return, $p
is still empty! This can be fixed as follows:
$p = array();
getArray($p);The following example works, as expected, in both 5.1.1 and 5.1.2.
<?
class A
{
public $i = 9;
}function doAmend(A &$a) // & not required because class instance
passed by reference by default
{
$a->i += 30;
}doAmend($b = new A); print_r($b); // this prints the value of member i as 39as expected.
?>
I'm guessing it has something to do with a memory leak. Either that, or
scope; creating a variable inside of a function call probably goes out
of scope after the function returns. Enable error_reporting(E_ALL|
E_STRICT) and see what happens.
--
Matt Sicker
Hi,
getArray($p = array());
Here you are relying on undefined behavior. It is not defined wether $p =
array() or the function call getarray($p) should be executed first so the
order might always change.
You should always use the two lines
$p = array();
getArray($p);
to be safe.
johannes
It looks like
http://www.php.net/manual/en/language.operators.assignment.php
could use a bit of updating. A paragraph about php 5 or a
link to the semantics changes for php 5 would help newbies.
The text about php4 says there is copy on assignment, which
implies a performance hit for using = instead of =&, but I
believe "copy on change" is more accurate. It wouldn't hurt
to explicitly say that "=" is generally better than "=&"
(unless you really need an alias). A user comment has been
there unchallenged for over 2 years saying you can pass an
assignment as a reference to a function, so it isn't surprising
that someone considered the 5.1.1 to 5.1.2 change an issue.
(Not that a user comment should bind the php devs, but when
the comment goes unchallenged, it can lead to false impressions.)
Perhaps some text explaining the assignment returns an r-value
would be helpful. (If people understand "($x = 1) = 2;"
shouldn't work, they might understand that passing "$x=1" to
a function by reference shouldn't affect $x. (or should
"($x = 1) = 2;" work and 5.1.2 had a real bug?))
I have to say the documentation for php is completely awesome
and is a major factor in our company using the language. I
just point out the above to help; it isn't a complaint.
BTW, I disagree with the post below that indicates the issue is
the order of calling. There is no way to call a function
before doing the work in the argument list. If I call f(1+2),
I can be sure the computation to get 3 will be performed
before calling f. The issue with f($p=array()) is that
"$p=array()" is an r-value.
- Todd
Hi,
getArray($p = array());Here you are relying on undefined behavior. It is not defined wether $p =
array() or the function call getarray($p) should be executed first so the
order might always change.You should always use the two lines
$p = array();
getArray($p);
to be safe.johannes
--
Todd Ruth <truth@proposaltech.com
Actually Todd, ...
There is no way to call a function
before doing the work in the argument list. If I call f(1+2),
I can be sure the computation to get 3 will be performed
before calling f. The issue with f($p=array()) is that
"$p=array()" is an r-value.
...that is theoretically very possible, and it's something called "lazy
evaluation": http://en.wikipedia.org/wiki/Lazy_evaluation
- Ron
"Todd Ruth" truth@proposaltech.com wrote in message
news:1153763607.30382.184.camel@notebook.local...
It looks like
http://www.php.net/manual/en/language.operators.assignment.php
could use a bit of updating. A paragraph about php 5 or a
link to the semantics changes for php 5 would help newbies.
The text about php4 says there is copy on assignment, which
implies a performance hit for using = instead of =&, but I
believe "copy on change" is more accurate. It wouldn't hurt
to explicitly say that "=" is generally better than "=&"
(unless you really need an alias). A user comment has been
there unchallenged for over 2 years saying you can pass an
assignment as a reference to a function, so it isn't surprising
that someone considered the 5.1.1 to 5.1.2 change an issue.
(Not that a user comment should bind the php devs, but when
the comment goes unchallenged, it can lead to false impressions.)
Perhaps some text explaining the assignment returns an r-value
would be helpful. (If people understand "($x = 1) = 2;"
shouldn't work, they might understand that passing "$x=1" to
a function by reference shouldn't affect $x. (or should
"($x = 1) = 2;" work and 5.1.2 had a real bug?))I have to say the documentation for php is completely awesome
and is a major factor in our company using the language. I
just point out the above to help; it isn't a complaint.BTW, I disagree with the post below that indicates the issue is
the order of calling. There is no way to call a function
before doing the work in the argument list. If I call f(1+2),
I can be sure the computation to get 3 will be performed
before calling f. The issue with f($p=array()) is that
"$p=array()" is an r-value.
- Todd
Hi,
getArray($p = array());Here you are relying on undefined behavior. It is not defined wether $p
=
array() or the function call getarray($p) should be executed first so
the
order might always change.You should always use the two lines
$p = array();
getArray($p);
to be safe.johannes
--
Todd Ruth <truth@proposaltech.com
in that case, the documentation on socket_select() needs to be edited:
http://www.php.net/socket-select says:
<?php
/* Prepare the read array */
$read = array($socket1, $socket2);
$num_changed_sockets = socket_select($read, $write = NULL, $except = NULL,
0);
if ($num_changed_sockets === false) {
/* Error handling /
} else if ($num_changed_sockets > 0) {
/ At least at one of the sockets something interesting happened */
}
?>
according to you, $write and $except could be undefined, right?
- Ron
"Johannes Schlueter" johannes@php.net wrote in message
news:200607241916.27024.johannes@php.net...
Hi,
getArray($p = array());Here you are relying on undefined behavior. It is not defined wether $p =
array() or the function call getarray($p) should be executed first so the
order might always change.You should always use the two lines
$p = array();
getArray($p);
to be safe.johannes