Doc #74257 [Opn->Fbk]: "namespace \X" does not work to declare a namespace
Edit report at https://bugs.php.net/bug.php?id=74257&edit=1
ID: 74257
Updated by: cmb@php.net
Reported by: david at grudl dot com
Summary: "namespace \X" does not work to declare a namespace
-Status: Open
+Status: Feedback
Type: Documentation Problem
Package: Scripting Engine problem
PHP Version: 7.1.2
-Assigned To:
+Assigned To: cmb
Block user comment: N
Private report: N
New Comment:
What is supposed to be missing from the documentation?
Previous Comments:
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[2019-01-31 22:51:34] roy-orbison at devo dot net dot au
Indeed, but that error is not picked up by linting, and the message is for an undefined constant.
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[2019-01-31 01:30:42] requinix@php.net
Keep in mind OP had to define X as a constant. In the real world namespaces have more meaningful
names and it's unlikely there will be a constant by the same name, so code would create an
error.
Additionally, "namespace\" is documented in "namespace keyword and __NAMESPACE__
constant"
http://php.net/manual/en/language.namespaces.nsconstants.php
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[2019-01-31 01:16:25] roy-orbison at devo dot net dot au
This is definitely a bug. If PHP devs really think it's a feature, why isn't it documented
in "Defining Namespaces"? There is no way anyone would expect the following to be anything
other than a definition.
namespace \Foo\Bar;
Accessing members of a namespace relative to global is done with a leading backslash, so why not the
same for defining? It should at least be a syntax error rather than evaluate and cause a fatal error
or silently fail. That type of expression could only read a constant without doing any assignment,
thus have no effect on other userspace state, right? So what's the point of allowing it without
even a warning?
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[2017-03-16 16:46:49] tpunt@php.net
^ See above comment...
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[2017-03-16 16:46:05] tpunt@php.net
This is not a bug, but rather a feature in namespaces known as relative names.
In short, the following line:
namespace \X;
is not a namespace *declaration*, but rather an *expression* that resolves to the symbol X of the
current namespace. The "namespace \" part (where space is optional) is either evaluated to
the currently defined namespace (it is simply ignored for the global namespace).
When utilising this feature, symbols that are not found in the current namespace will not fallback
to the globally defined functions or constants.
So, the following works:
var_dump(namespace\PHP_ZTS);
But the following will error (due to an undefined PHP_ZTS constant in the current namespace):
namespace test;
var_dump(namespace\PHP_ZTS);
Whilst obviously the following will work as expected (by falling back to the globally defined
PHP_ZTS constant):
namespace test;
var_dump(PHP_ZTS);
See the manual for further information: http://php.net/manual/en/language.namespaces.rules.php
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