Edit report at https://bugs.php.net/bug.php?id=54740&edit=1
ID: 54740
Comment by: marrch dot caat at gmail dot com
Reported by: dukeofgaming at gmail dot com
Summary: Ternary operator will not work with return by
reference
Status: Not a bug
Type: Bug
Package: Scripting Engine problem
PHP Version: Irrelevant
Block user comment: N
Private report: N
New Comment:
Thank you, Nikic! You've removed scales from my eyes. To my great pity and
shame, I didn't understand that through all my working experience and really
though & is a "take reference" operator, as one as exists in C/C++ etc :( Thank
you once again for your explanation! Now I see this is not really a bug...
Previous Comments:
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[2013-10-02 11:20:20] [email protected]
@marrch: PHP has no concept of a general & operator that takes the "reference"
of a variable(whatever that's supposed to be). PHP only has a =& operator
(which is really "=&" and not a combination of "=" and "&") which expects a
variable on the right hand side. PHP also has a & modifier for function
arguments and return values.
If you want to do yourself and others a favor, write $foo =& $bar rather than
$foo = &$bar to make sure that there are no misunderstandings regarding this ;)
------------------------------------------------------------------------
[2013-10-02 11:12:00] marrch dot caat at gmail dot com
Mike, I understand that. The second note tells I caanot return a reference to
an expression result, such as &$object->method() or &(new StdClass()) - I can
understand that. But the code sample I provided doesn't try to do that. To make
things even simplier, the following code still fails to compile:
$link = $flag ? &$a : &$b;
It doesn't try to return a reference to an expression, just a reference to a
viriable; It doesn't try doing anything that the following code doesn't:
if ($flag)
$link = &$a;
else
$link = &$b;
And maybi I'm really stupid, but after 10 years in PHP development I still
don't understand why the first code cannot be compiled :(
------------------------------------------------------------------------
[2013-10-02 05:27:05] [email protected]
I meant the documentation "Note:" (warning) not the user-contributed note.
------------------------------------------------------------------------
[2013-10-01 20:35:33] marrch dot caat at gmail dot com
I thoroughly read the article you mentioned, Mike, but still don't understand
why the following code fails to compile:
$link = isset($i) ? (& $arr[$i]) : null;
- while the following works fine:
$link = &$arr[$i];
In this case, &$arr[$i] is a legal reference assignment, so the first code
should behave equal to
if (isset($i)) {
$link = &$arr[$i];
} else {
$link = null;
}
- but this code works fine, and mentioned above isn't even compiled. What's
wrong with it?
------------------------------------------------------------------------
[2013-10-01 14:42:02] [email protected]
Thank you for taking the time to write to us, but this is not
a bug. Please double-check the documentation available at
http://www.php.net/manual/ and the instructions on how to report
a bug at http://bugs.php.net/how-to-report.php
Check the second note here:
http://php.net/manual/en/language.references.return.php
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Edit this bug report at https://bugs.php.net/bug.php?id=54740&edit=1
