Bug #54740 [Com]: Ternary operator will not work with return by reference
Edit report at https://bugs.php.net/bug.php?id=54740edit=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:
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 :(
Previous Comments:
[2013-10-02 05:27:05] m...@php.net
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] m...@php.net
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
[2012-08-27 14:17:44] marrch dot caat at gmail dot com
This is a general problem with reference inside ternary operator. For ex., the
following script fails with the same error:
$link = isset($i) ? ( $arr[$i]) : null;
- while the following works fine:
$link = $arr[$i];
[2011-05-15 22:59:17] dukeofgaming at gmail dot com
Description:
PHP fails to parse a returned by reference value when using the ternary
operator.
The test script provided illustrates a case of when it is absolutely necessary
to return by reference; if the is removed then the output would be a fatal
error: Fatal error: Cannot use [] for reading in ...
Test script:
---
$value = ($condition)?(
$some_value
):($object-Collection[]);
Expected result:
No errors, should be the equivalent of having:
if($condition){
$value = $some_value;
}else{
$value = $object-Collection[];
}
Actual result:
--
Parse error: syntax error, unexpected '' in ...
--
Edit this bug report at https://bugs.php.net/bug.php?id=54740edit=1
Bug #54740 [Com]: Ternary operator will not work with return by reference
Edit report at https://bugs.php.net/bug.php?id=54740edit=1
ID: 54740
Comment by: ni...@php.net
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:
@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 ;)
Previous Comments:
[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] m...@php.net
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] m...@php.net
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
[2012-08-27 14:17:44] marrch dot caat at gmail dot com
This is a general problem with reference inside ternary operator. For ex., the
following script fails with the same error:
$link = isset($i) ? ( $arr[$i]) : null;
- while the following works fine:
$link = $arr[$i];
The remainder of the comments for this report are too long. To view
the rest of the comments, please view the bug report online at
https://bugs.php.net/bug.php?id=54740
--
Edit this bug report at https://bugs.php.net/bug.php?id=54740edit=1
Bug #54740 [Com]: Ternary operator will not work with return by reference
Edit report at https://bugs.php.net/bug.php?id=54740edit=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:
[2013-10-02 11:20:20] ni...@php.net
@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] m...@php.net
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] m...@php.net
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
The remainder of the comments for this report are too long. To view
the rest of the comments, please view the bug report online at
https://bugs.php.net/bug.php?id=54740
--
Edit this bug report at https://bugs.php.net/bug.php?id=54740edit=1
Bug #54740 [Com]: Ternary operator will not work with return by reference
Edit report at https://bugs.php.net/bug.php?id=54740edit=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:
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?
Previous Comments:
[2013-10-01 14:42:02] m...@php.net
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
[2012-08-27 14:17:44] marrch dot caat at gmail dot com
This is a general problem with reference inside ternary operator. For ex., the
following script fails with the same error:
$link = isset($i) ? ( $arr[$i]) : null;
- while the following works fine:
$link = $arr[$i];
[2011-05-15 22:59:17] dukeofgaming at gmail dot com
Description:
PHP fails to parse a returned by reference value when using the ternary
operator.
The test script provided illustrates a case of when it is absolutely necessary
to return by reference; if the is removed then the output would be a fatal
error: Fatal error: Cannot use [] for reading in ...
Test script:
---
$value = ($condition)?(
$some_value
):($object-Collection[]);
Expected result:
No errors, should be the equivalent of having:
if($condition){
$value = $some_value;
}else{
$value = $object-Collection[];
}
Actual result:
--
Parse error: syntax error, unexpected '' in ...
--
Edit this bug report at https://bugs.php.net/bug.php?id=54740edit=1
Bug #54740 [Com]: Ternary operator will not work with return by reference
Edit report at https://bugs.php.net/bug.php?id=54740edit=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: Open
Type: Bug
Package:Compile Failure
PHP Version:Irrelevant
Block user comment: N
Private report: N
New Comment:
This is a general problem with reference inside ternary operator. For ex., the
following script fails with the same error:
$link = isset($i) ? ( $arr[$i]) : null;
- while the following works fine:
$link = $arr[$i];
Previous Comments:
[2011-05-15 22:59:17] dukeofgaming at gmail dot com
Description:
PHP fails to parse a returned by reference value when using the ternary
operator.
The test script provided illustrates a case of when it is absolutely necessary
to return by reference; if the is removed then the output would be a fatal
error: Fatal error: Cannot use [] for reading in ...
Test script:
---
$value = ($condition)?(
$some_value
):($object-Collection[]);
Expected result:
No errors, should be the equivalent of having:
if($condition){
$value = $some_value;
}else{
$value = $object-Collection[];
}
Actual result:
--
Parse error: syntax error, unexpected '' in ...
--
Edit this bug report at https://bugs.php.net/bug.php?id=54740edit=1
