Edit report at https://bugs.php.net/bug.php?id=51225&edit=1
ID: 51225
User updated by: xiaodujinjin at gmail dot com
Reported by: xiaodujinjin at gmail dot com
Summary: cannot define a class with the same name as an
interface
Status: Assigned
Type: Bug
Package: Class/Object related
Operating System: Windows XP MT15i
PHP Version: 5.2.13
Assigned To: johannes
Block user comment: N
Private report: N
New Comment:
xiaodujinjin@gmail.com
Previous Comments:
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[2012-05-20 02:39:55] xiaodujinjin at gmail dot com
xiaodujinjin@gmail.com
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[2012-03-20 21:05:19] LaKing at D250 dot hu
The language-developers have two options. Either a clearer error message as a quick-fix, or allowing
interfaces and classes with the same name, with the necessary underlying updates properly
implemented in the language. That would be the elegant, and logical way.
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[2010-09-23 08:03:40] i at walkinraven dot name
I think you the fact of class and interface sharing a same namespace should be clearly written into
the manual!
And the better way, the two should not share a same namespace at all.
It waste me some much time to check my project.
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[2010-08-05 23:38:58] bobalicious dot bob at gmail dot com
In my opinion, in order to be valid the code snippet should read:
if (!class_exists('foobar') && !interface_exists('foobar') ) {
class foobar{}
}
The error message on attempting to declare a class with the same name as an
interface should respond:
Cannot declare class as an interface exists with that name
The reverse message should also be possible.
It does not make sense to allow an interface and a class to have the same name
(type hinting is a great example why not), and 'class_exists' should only refer
to classes (the clue's in the name).
The confusion is merely down to an inaccurate error message.
Note: There's a reason it's not uncommon for people to prefix interfaces with an
'i'
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[2010-03-08 10:35:06] tony at marston-home dot demon dot co dot uk
I disagree. class_exists() SHOULD check if that name has already been declared as an interface
otherwise you get the following situation:
if (!class_exists('foobar') { // returns false
class foobar{} // fails because interface exists
}
On the one hand it is saying "a class with the name 'foobar' does not exist"
which is immediately followed by "you cannot create a class with the name 'foobar' as
it already exists". That is not logical to me.
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