autoload and namespaces

Question

I've been working with PHP for a long time, but am now starting to experiment with newer language features such as namespaces. I have a question regarding autoloading that I haven't been able to find an adequate answer to in my web searching.

Suppose I have classes in different namespaces:

namespace foo\bar\baz;

class Quux
{
}

namespace fred\barney\wilma;

class Betty
{
}

Then suppose I had an autoloader that assumes that there's a 1:1 mapping between namespaces and directory structures:

function autoload ($className)
{
    $className = str_replace ('\\', DIRECTORY_SEPERATOR, $className);
    include ($className . 'php');
}

spl_autoload_register ('autoload');

Does the fully qualified namespace get passed to the autoloader under all circumstances, or does the autoloader need to take the namespace currently being used into account?

For example, if I do the following:

$a = new \foo\bar\baz\Quux;
$b = new \fred\barney\wilma\Betty;

the autoloader should work fine.

But what if I do the following?

use \foo\bar\baz as FBB;
$a = new Quux;
$b = new \fred\barney\wilma\Betty;

When attempting to instantiate a new Quux, will the autoloader still get \foo\bar\baz\Quux as the class name argument? Or should it get FBB\Quux, or even just Quux?

If the latter, can I determine the namespace the class is supposed to be in from within my autoloader by using __NAMESPACE__ or some other such mechanism?

So what is the result you have tried? – xdazz Sep 18 '11 at 10:13 — xdazz, Sep 18 '11 at 10:13

score 4 · Accepted Answer · answered Sep 18 '11 at 10:44

4

The autoloader will get foo\bar\baz\Quux as the class name argument.

Namespace name definitions
Unqualified name
This is an identifier without a namespace separator, such as Foo

Qualified name
This is an identifier with a namespace separator, such as Foo\Bar

Fully qualified name
This is an identifier with a namespace separator that begins with a namespace separator, such as \Foo\Bar. namespace\Foo is also a fully qualified name.

And the the rule is :

All unqualified and qualified names (not fully qualified names) are translated during compilation according to current import rules. For example, if the namespace A\B\C is imported as C, a call to C\D\e() is translated to A\B\C\D\e().

answered Sep 18 '11 at 10:44

xdazz

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Sorry, but I had to unaccept this answer I'm afraid. When I tried actually implementing an autoloader that worked like this, I found that it worked okay for `new /foo/bar/baz ()`, but if I did `use /foo/bar; new baz()` I'd only get baz being passed to my autoloader. – GordonM Sep 21 '11 at 21:36
@GordonM You should do `use foo\bar\baz; new baz();`. – xdazz Sep 22 '11 at 01:42
1

Yeah, I just figured that out. Sorry about that, I'd just totally misunderstood how the use keyword worked. – GordonM Sep 22 '11 at 05:14

score 0 · Answer 2 · answered Sep 18 '11 at 11:28

You will get a fully qualified namespace name (without leading backslash).

This also applies to all other dynamic language constructs and functions in PHP:

If you write $obj = new $class the $class must be fully qualified and doesn't use any defined aliases.
If you write class_exists($class) the $class must also be fully qualified.
And same applies to autoload: It is passed the fully qualified name.

(The only exception to this rule I know of is the define function, which can define constants both using fully qualified names, but also using an unqualified name, in which case it will be defined in the current namespace.)

autoload and namespaces

2 Answers2

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