Bug #52031 [Bgs]: Passing 0 as a parameter through a custom method
Edit report at http://bugs.php.net/bug.php?id=52031edit=1
ID: 52031
User updated by: d dot reade at ReadesGroupServices dot com
Reported by: d dot reade at ReadesGroupServices dot com
Summary: Passing 0 as a parameter through a custom method
Status: Bogus
Type: Bug
Package: Unknown/Other Function
Operating System: CentOS 5.5
PHP Version: 5.3.2
New Comment:
Sorry but that makes no sense. How is this not a bug?
If I pass 0 as the parameter and the first if statement is empty(),
PHP knows the paramater is empty and the if statement is triggered.
Whereas without empty() PHP executes the first statement regardless even
though the same value would trigger empty()!
It shouldn't do that - if the first if statement is ($str == 'Test'),
PHP should skip it just as it would with any other string that doesn't
match.
Previous Comments:
[2010-06-09 14:51:46] degeb...@php.net
Sorry, but your problem does not imply a bug in PHP itself. For a
list of more appropriate places to ask for help using PHP, please
visit http://www.php.net/support.php as this bug system is not the
appropriate forum for asking support questions. Due to the volume
of reports we can not explain in detail here why your report is not
a bug. The support channels will be able to provide an explanation
for you.
Thank you for your interest in PHP.
See: http://php.net/manual/en/language.types.type-juggling.php
[2010-06-09 14:41:32] d dot reade at ReadesGroupServices dot com
Description:
Sorry but I didn't know how to best word the summary.
I have a custom method with an if statement inside. If the parameter
passed through matches Test, a string is echo'ed. However if I pass
0 through as the parameter, PHP thinks this is Test and echoes the
string.
Passing a blank parameter through, i.e. , does not have the same
effect as 0.
Inserting if (empty($str)) before the if statement in the example
works as expected, catches the 0 parameter and I get the expected
result. However without this, I get the actual result.
All plugins have been disabled and Apache restarted.
Test script:
---
?php
function testMethod($str = 0)
{
if ($str == 'Test')
{
echo 'Test String!br';
}
}
testMethod('Test');
testMethod(0);
testMethod();
testMethod(1);
testMethod('Something');
?
Expected result:
Test String!
Actual result:
--
Test String!
Test String!
Test String!
--
Edit this bug report at http://bugs.php.net/bug.php?id=52031edit=1
Bug #52031 [Bgs]: Passing 0 as a parameter through a custom method
Edit report at http://bugs.php.net/bug.php?id=52031edit=1
ID: 52031
Updated by: degeb...@php.net
Reported by: d dot reade at ReadesGroupServices dot com
Summary: Passing 0 as a parameter through a custom method
Status: Bogus
Type: Bug
Package: Unknown/Other Function
Operating System: CentOS 5.5
PHP Version: 5.3.2
New Comment:
You are comparing an integer with a string, so the string gets converted
to an int. Because (int)Test is (int)0 and 0==0, the condition
passes.
It's not a bug, and it's documented behavior. If you want it to check
the type as well, you need to use the === operator as documented.
Previous Comments:
[2010-06-09 14:57:52] d dot reade at ReadesGroupServices dot com
Sorry but that makes no sense. How is this not a bug?
If I pass 0 as the parameter and the first if statement is empty(),
PHP knows the paramater is empty and the if statement is triggered.
Whereas without empty() PHP executes the first statement regardless even
though the same value would trigger empty()!
It shouldn't do that - if the first if statement is ($str == 'Test'),
PHP should skip it just as it would with any other string that doesn't
match.
[2010-06-09 14:51:46] degeb...@php.net
Sorry, but your problem does not imply a bug in PHP itself. For a
list of more appropriate places to ask for help using PHP, please
visit http://www.php.net/support.php as this bug system is not the
appropriate forum for asking support questions. Due to the volume
of reports we can not explain in detail here why your report is not
a bug. The support channels will be able to provide an explanation
for you.
Thank you for your interest in PHP.
See: http://php.net/manual/en/language.types.type-juggling.php
[2010-06-09 14:41:32] d dot reade at ReadesGroupServices dot com
Description:
Sorry but I didn't know how to best word the summary.
I have a custom method with an if statement inside. If the parameter
passed through matches Test, a string is echo'ed. However if I pass
0 through as the parameter, PHP thinks this is Test and echoes the
string.
Passing a blank parameter through, i.e. , does not have the same
effect as 0.
Inserting if (empty($str)) before the if statement in the example
works as expected, catches the 0 parameter and I get the expected
result. However without this, I get the actual result.
All plugins have been disabled and Apache restarted.
Test script:
---
?php
function testMethod($str = 0)
{
if ($str == 'Test')
{
echo 'Test String!br';
}
}
testMethod('Test');
testMethod(0);
testMethod();
testMethod(1);
testMethod('Something');
?
Expected result:
Test String!
Actual result:
--
Test String!
Test String!
Test String!
--
Edit this bug report at http://bugs.php.net/bug.php?id=52031edit=1
Bug #52031 [Bgs]: Passing 0 as a parameter through a custom method
Edit report at http://bugs.php.net/bug.php?id=52031edit=1
ID: 52031
User updated by: d dot reade at ReadesGroupServices dot com
Reported by: d dot reade at ReadesGroupServices dot com
Summary: Passing 0 as a parameter through a custom method
Status: Bogus
Type: Bug
Package: Unknown/Other Function
Operating System: CentOS 5.5
PHP Version: 5.3.2
New Comment:
I see you what mean. Thanks for the clarification, much appreciated.
Previous Comments:
[2010-06-09 15:14:38] degeb...@php.net
You are comparing an integer with a string, so the string gets converted
to an int. Because (int)Test is (int)0 and 0==0, the condition
passes.
It's not a bug, and it's documented behavior. If you want it to check
the type as well, you need to use the === operator as documented.
[2010-06-09 14:57:52] d dot reade at ReadesGroupServices dot com
Sorry but that makes no sense. How is this not a bug?
If I pass 0 as the parameter and the first if statement is empty(),
PHP knows the paramater is empty and the if statement is triggered.
Whereas without empty() PHP executes the first statement regardless even
though the same value would trigger empty()!
It shouldn't do that - if the first if statement is ($str == 'Test'),
PHP should skip it just as it would with any other string that doesn't
match.
[2010-06-09 14:51:46] degeb...@php.net
Sorry, but your problem does not imply a bug in PHP itself. For a
list of more appropriate places to ask for help using PHP, please
visit http://www.php.net/support.php as this bug system is not the
appropriate forum for asking support questions. Due to the volume
of reports we can not explain in detail here why your report is not
a bug. The support channels will be able to provide an explanation
for you.
Thank you for your interest in PHP.
See: http://php.net/manual/en/language.types.type-juggling.php
[2010-06-09 14:41:32] d dot reade at ReadesGroupServices dot com
Description:
Sorry but I didn't know how to best word the summary.
I have a custom method with an if statement inside. If the parameter
passed through matches Test, a string is echo'ed. However if I pass
0 through as the parameter, PHP thinks this is Test and echoes the
string.
Passing a blank parameter through, i.e. , does not have the same
effect as 0.
Inserting if (empty($str)) before the if statement in the example
works as expected, catches the 0 parameter and I get the expected
result. However without this, I get the actual result.
All plugins have been disabled and Apache restarted.
Test script:
---
?php
function testMethod($str = 0)
{
if ($str == 'Test')
{
echo 'Test String!br';
}
}
testMethod('Test');
testMethod(0);
testMethod();
testMethod(1);
testMethod('Something');
?
Expected result:
Test String!
Actual result:
--
Test String!
Test String!
Test String!
--
Edit this bug report at http://bugs.php.net/bug.php?id=52031edit=1
