Edit report at https://bugs.php.net/bug.php?id=64755&edit=1
ID: 64755
Updated by: ni...@php.net
Reported by: eth at ethaniel dot com
Summary: Only variables should be passed by reference
Status: Not a bug
Type: Bug
Package: Scripting Engine problem
Operating System: Debian 7.0
PHP Version: 5.4.14
Block user comment: N
Private report: N
New Comment:
array_shift(array_splice($dbents, $x, 1)) does not throw an ERROR exception
(whatever that is). It throws an E_STRICT level error, which is the lowest
error type we have.
The E_STRICT tells you that you have some code that currently works, but is
discouraged. What you were actually looking for is array_splice($dbents, $x,
1)[0] (at least I assume so, because I don't really get your code sample.)
Previous Comments:
------------------------------------------------------------------------
[2013-09-11 23:13:24] guy at syntheticwebapps dot com
Even worse, the following no longer works in PHP 5.4+ and this is an ERROR
exception:
$dbentry = array_shift(array_splice($dbents, $x, 1)))
This is why some leeway is appropriate on this error. I you can't chain
together
such logical progressions without INVENTING a variable in the code which has no
other purpose than to avoid this ERROR exception, the language is headed the
wrong direction and causing limitations for no cause.
I really think this needs to be fixed.
------------------------------------------------------------------------
[2013-09-11 22:59:26] guy at syntheticwebapps dot com
I have a different example than the one given. Consider:
function myfunc( &$refvar ) {
...
}
$horse = myfunc( $value = array() );
The code example above also gets the same exception, however, this does not
seem
correct. This suggests that the type of "$value = array()" is expression and
not
a variable. But this violates an assumption in the language. Is this assumption
incorrect? Should it be incorrect?
As in the original reported form of the problem, I believe there should be some
forgiveness in the call-by-reference receiver. I can understand why NOT to do
that, but there are safe ways to do it anyway, like create a new zval for local
use.
But it's more important to me that $value = array() evaluates to the VARIABLE
$value and NOT an expression. What expression could it possibly represent? The
supposed expression in such a case is meaningless, whereas the variable is
meaningful and useful. Any chance this could be fixed in future?
------------------------------------------------------------------------
[2013-05-02 08:25:11] paj...@php.net
See www.php.net/array_pop.
Using: array_pop(array_keys($a)); array_keys($a) is an expression while
array_pop
expects a variable. ($k = array_keys($a);...).
------------------------------------------------------------------------
[2013-05-02 08:03:55] eth at ethaniel dot com
Description:
------------
I get a "PHP Strict Standards: Only variables should be passed by reference
in"
error where there should be none.
Test script:
---------------
echo array_pop(array_keys(array("erwre")));
Expected result:
----------------
Result: 0.
Actual result:
--------------
The script returns the result (0), but also a "PHP Strict Standards: Only
variables should be passed by reference in" in my error log.
------------------------------------------------------------------------
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Edit this bug report at https://bugs.php.net/bug.php?id=64755&edit=1
