forbid use declaration outside of a namespace in PHP 7

11 years ago by Robert Stoll — view source — reply

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Heya,

I always found it very ugly that it is possible to define a use outside of a namespace. Consider the following:

namespace{ //default namespace
}

use foo\Bar;

namespace test{
new Bar(); //error, test\Bar not found
}

After some thoughts it is quite clear that Bar is test\Bar and not foo\Bar inside of the namespace test. But consider
the following example which is not so obvious:

use foo\Bar;
namespace test;
new Bar(); //error, test\Bar not found

The use declaration looks like a normal use declaration at first glance.
I do not see why we should actually support this "feature" any longer and thus suggest to remove it in PHP 7. Although,
it is not a bug (the use declaration is simply ignored as far as I can tell) I suppose it confuses the user.
Nevertheless, even if we declare it as a "feature" I think it should at least not be a "feature" of the specification of
PHP 7.

Thoughts?

Cheers,
Robert

ps: I first started the discussion @standards, just if you should wonder why it pops up here now as well:
http://news.php.net/php.standards/528

11 years ago by Robert Stoll — view source — reply

unread

-----Ursprüngliche Nachricht-----
Von: Robert Stoll [mailto:php@tutteli.ch]
Gesendet: Mittwoch, 29. Oktober 2014 20:55
An: 'PHP Internals'
Betreff: [PHP-DEV] forbid use declaration outside of a namespace in PHP 7

Heya,

I always found it very ugly that it is possible to define a use outside of a namespace. Consider the following:

namespace{ //default namespace
}

use foo\Bar;

namespace test{
new Bar(); //error, test\Bar not found }

After some thoughts it is quite clear that Bar is test\Bar and not foo\Bar inside of the namespace test. But consider
the
following example which is not so obvious:

use foo\Bar;
namespace test;
new Bar(); //error, test\Bar not found

The use declaration looks like a normal use declaration at first glance.
I do not see why we should actually support this "feature" any longer and thus suggest to remove it in PHP 7.
Although, it is
not a bug (the use declaration is simply ignored as far as I can tell) I suppose it confuses the user.
Nevertheless, even if we declare it as a "feature" I think it should at least not be a "feature" of the specification
of PHP 7.

Thoughts?

Cheers,
Robert

ps: I first started the discussion @standards, just if you should wonder why it pops up here now as well:
http://news.php.net/php.standards/528

--

No one else who thinks this "feature" should be removed in PHP 7 ?

11 years ago by Chris Wright — view source — reply

unread

-----Ursprüngliche Nachricht-----
Von: Robert Stoll [mailto:php@tutteli.ch]
Gesendet: Mittwoch, 29. Oktober 2014 20:55
An: 'PHP Internals'
Betreff: [PHP-DEV] forbid use declaration outside of a namespace in PHP 7

Heya,

I always found it very ugly that it is possible to define a use outside
of a namespace. Consider the following:

namespace{ //default namespace
}

use foo\Bar;

namespace test{
new Bar(); //error, test\Bar not found }

After some thoughts it is quite clear that Bar is test\Bar and not
foo\Bar inside of the namespace test. But consider
the
following example which is not so obvious:

use foo\Bar;
namespace test;
new Bar(); //error, test\Bar not found

The use declaration looks like a normal use declaration at first glance.
I do not see why we should actually support this "feature" any longer
and thus suggest to remove it in PHP 7.
Although, it is
not a bug (the use declaration is simply ignored as far as I can tell) I
suppose it confuses the user.
Nevertheless, even if we declare it as a "feature" I think it should at
least not be a "feature" of the specification
of PHP 7.

Thoughts?

Cheers,
Robert

ps: I first started the discussion @standards, just if you should wonder
why it pops up here now as well:
http://news.php.net/php.standards/528

--
To unsubscribe,
visit: http://www.php.net/unsub.php

No one else who thinks this "feature" should be removed in PHP 7 ?

I would say that the lack of anyone saying "no, don't do this" probably
means everyone is OK with it. Suggest you work up a patch and PR it, then
ping the list to highlight it for further discussion.

11 years ago by Robert Stoll — view source — reply

unread

I would say that the lack of anyone saying "no, don't do this" probably means everyone is OK with it. Suggest you work up a
patch and PR it, then ping the list to highlight it for further discussion.

Sounds reasonable but unfortunately I do not have time at the moment to work up a patch and when I have time (in half a year or so) it might be too late for PHP 7 and the change would need to wait for PHP 8 since it is not backward compatible. That is the reason why I hope we could agree on the topic, remove the corresponding sentence from the spec and I could patch up a fix when I have time for it.

11 years ago by Andrea Faulds — view source — reply

unread

I would say that the lack of anyone saying "no, don't do this" probably means everyone is OK with it. Suggest you work up a
patch and PR it, then ping the list to highlight it for further discussion.

Sounds reasonable but unfortunately I do not have time at the moment to work up a patch and when I have time (in half a year or so) it might be too late for PHP 7 and the change would need to wait for PHP 8 since it is not backward compatible. That is the reason why I hope we could agree on the topic, remove the corresponding sentence from the spec and I could patch up a fix when I have time for it.

I don’t like the sound of changing the spec before the implementation.

Andrea Faulds
http://ajf.me/

11 years ago by Stas Malyshev — view source — reply

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Hi!

I don’t like the sound of changing the spec before the implementation.

I agree - whatever are our considerations and limitations on what we can
and can not do in PHP 7, the spec should reflect what is there.
Otherwise we end up releasing an implementation that does not match our
own spec, that'd be embarrassing.

--
Stanislav Malyshev, Software Architect
SugarCRM: http://www.sugarcrm.com/

11 years ago by Robert Stoll — view source — reply

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-----Ursprüngliche Nachricht-----
Von: Stas Malyshev [mailto:smalyshev@sugarcrm.com]
Gesendet: Mittwoch, 12. November 2014 00:45
An: Andrea Faulds; Robert Stoll
Cc: Chris Wright; PHP Internals
Betreff: Re: [PHP-DEV] forbid use declaration outside of a namespace in PHP 7

Hi!

I don’t like the sound of changing the spec before the implementation.

I agree - whatever are our considerations and limitations on what we can and can not do in PHP 7, the spec should reflect
what is there.
Otherwise we end up releasing an implementation that does not match our own spec, that'd be embarrassing.

Alright, fair enough. I will ping this list once a patch is ready.

11 years ago by Adam Harvey — view source — reply

unread

I always found it very ugly that it is possible to define a use outside of a namespace. Consider the following:

namespace{ //default namespace
}

use foo\Bar;

namespace test{
new Bar(); //error, test\Bar not found }

After some thoughts it is quite clear that Bar is test\Bar and not foo\Bar inside of the namespace test. But consider
the
following example which is not so obvious:

use foo\Bar;
namespace test;
new Bar(); //error, test\Bar not found

The use declaration looks like a normal use declaration at first glance.
I do not see why we should actually support this "feature" any longer and thus suggest to remove it in PHP 7.
Although, it is
not a bug (the use declaration is simply ignored as far as I can tell) I suppose it confuses the user.
Nevertheless, even if we declare it as a "feature" I think it should at least not be a "feature" of the specification
of PHP 7.

Sorry, I apparently missed this the first time. Would this mean that
this sort of script (where we're using use statements without a
namespace) would no longer work?

<?php
use GuzzleHttp\Client;

include DIR.'/vendor/autoload.php';

$client = new Client;
// $client is a GuzzleHttp\Client object
?>

Because that strikes me as an irritating and unnecessary BC break for
people who are writing throwaway scripts (with no need to live in
namespaces) that pull in namespaced libraries to me.

Adam

11 years ago by Robert Stoll — view source — reply

unread

Sorry, I apparently missed this the first time. Would this mean that this sort of script (where we're using use statements
without a
namespace) would no longer work?

<?php
use GuzzleHttp\Client;

include DIR.'/vendor/autoload.php';

$client = new Client;
// $client is a GuzzleHttp\Client object ?>

Because that strikes me as an irritating and unnecessary BC break for people who are writing throwaway scripts (with no
need to live in
namespaces) that pull in namespaced libraries to me.

Adam

That's still perfectly fine because in your code the use statement is not outside of a namespace, it is implicitly in the default namespace.
I am talking about the two following scenarios:

use \Exception;
namespace test;

$e = new Exception(); //would fails because test\Exception is unknown -> use was defined outside the namespace

Or
namespace a{
}

use some\UseDeclaration\which\is\outside\of\AnyNamespace as Useless; // sorry for begin biased ^^

namespace test{
}

Cheers,
Robert

11 years ago by Rowan Collins — view source — reply

unread

Robert Stoll wrote on 12/11/2014 21:27:

That's still perfectly fine because in your code the use statement is not outside of a namespace, it is implicitly in the default namespace.

Surely if that's true of Adam's example, it's true of yours as well?

namespace a{
}

use some\UseDeclaration\which\is\outside\of\AnyNamespace as Useless; // sorry for begin biased ^^

namespace test{
}

This scenario looks identical to Adam's to me, unless you propose to
look at the entirety of the file to detect if there is something other
than a use statement in the default namespace.

For that matter, what about this:

use GuzzleHttp\Client;

include DIR.'/vendor/autoload.php';

$client = new Client;
// $client is a GuzzleHttp\Client object

namespace Foo;
$client = new Client;
// $client is a Foo\Client object

Certainly this code is not sensible, but it does have a meaningful
behaviour, which derives directly from the previous examples and
documented uses of the namespace keyword. I'm not sure how you'd
distinguish between the various combinations here.

Regards,

Rowan Collins
[IMSoP]

11 years ago by Robert Stoll — view source — reply

unread

-----Ursprüngliche Nachricht-----
Von: Rowan Collins [mailto:rowan.collins@gmail.com]
Gesendet: Donnerstag, 13. November 2014 11:57
An: internals@lists.php.net
Betreff: Re: AW: [PHP-DEV] forbid use declaration outside of a namespace in PHP 7

Robert Stoll wrote on 12/11/2014 21:27:

That's still perfectly fine because in your code the use statement is not outside of a namespace, it is implicitly in the
default namespace.

Surely if that's true of Adam's example, it's true of yours as well?

[Robert Stoll]
The difference is that I have specified a namespace and thus code outside of that namespace is no longer implicitly in the default namespace.

namespace a{
}

use some\UseDeclaration\which\is\outside\of\AnyNamespace as Useless;
// sorry for begin biased ^^

namespace test{
}

This scenario looks identical to Adam's to me, unless you propose to look at the entirety of the file to detect if there is
something other than a use statement in the default namespace.

[Robert Stoll]
Same applies here, my use statement is outside of a namespace Adam's isn't.

For that matter, what about this:

use GuzzleHttp\Client;

include DIR.'/vendor/autoload.php';

$client = new Client;
// $client is a GuzzleHttp\Client object

namespace Foo;
$client = new Client;
// $client is a Foo\Client object

Certainly this code is not sensible, but it does have a meaningful behaviour, which derives directly from the previous
examples and documented uses of the namespace keyword. I'm not sure how you'd distinguish between the various
combinations here.

Regards,

Rowan Collins
[IMSoP]

[Robert Stoll]
I haven't tested it I am pretty sure this script would fail since $client = new Client is outside of a namespace - an error message something like "namespace must be the first statement" would show up.
If I rewrite your code to the following:

namespace{
use GuzzleHttp\Client;

include DIR.'/vendor/autoload.php';

$client = new Client;
// $client is a GuzzleHttp\Client object

namespace Foo{
$client = new Client;
// $client is a Foo\Client object
}

Then it would make sense (use statement is inside a namespace).

11 years ago by Rowan Collins — view source — reply

unread

Robert Stoll wrote on 13/11/2014 12:30:

-----Ursprüngliche Nachricht-----

Von: Rowan Collins [mailto:rowan.collins@gmail.com]
Gesendet: Donnerstag, 13. November 2014 11:57
An:internals@lists.php.net
Betreff: Re: AW: [PHP-DEV] forbid use declaration outside of a namespace in PHP 7

Robert Stoll wrote on 12/11/2014 21:27:

That's still perfectly fine because in your code the use statement is not outside of a namespace, it is implicitly in the
default namespace.

Surely if that's true of Adam's example, it's true of yours as well?
[Robert Stoll]
The difference is that I have specified a namespace and thus code outside of that namespace is no longer implicitly in the default namespace.

Ah, OK, on further experimenting, and closer reading of the
documentation, I see that you are right.

For anyone else wondering, the key quotes from the manual are:

"A file containing a namespace must declare the namespace at the top
of the file before any other code - with one exception: the declare
http://php.net/manual/en/control-structures.declare.php keyword."
"No PHP code may exist outside of the namespace brackets except for an
opening declare statement."

So as soon as the keyword "namespace" appears somewhere in the file,
there is no longer any implicit global non-namespaced code.

As such, it seems to me that allowing a "use" statement is simply a bug

no other code is allowed in that position, and the only documented
exception to that rule is declare().

Regards,

Rowan Collins
[IMSoP]

11 years ago by johannes@schlueters.de — view source — reply

unread

That's still perfectly fine because in your code the use statement is not outside of a namespace, it is implicitly in the default namespace.
I am talking about the two following scenarios:

use \Exception;
namespace test;

$e = new Exception(); //would fails because test\Exception is unknown -> use was defined outside the namespace

Or
namespace a{
}

use some\UseDeclaration\which\is\outside\of\AnyNamespace as Useless; // sorry for begin biased ^^

namespace test{
}

If I get it rigth this might break code by people who, for whatever
reason, combine multiple PHP files into a single one. (cat *.php >
full.php && php full.php)

johannes

11 years ago by Chris Wright — view source — reply

unread

That's still perfectly fine because in your code the use statement is
not outside of a namespace, it is implicitly in the default namespace.
I am talking about the two following scenarios:

use \Exception;
namespace test;

$e = new Exception(); //would fails because test\Exception is unknown ->
use was defined outside the namespace

Or
namespace a{
}

use some\UseDeclaration\which\is\outside\of\AnyNamespace as Useless; //
sorry for begin biased ^^

namespace test{
}

If I get it rigth this might break code by people who, for whatever
reason, combine multiple PHP files into a single one. (cat *.php >
full.php && php full.php)

This would only break if the individual files were already invalid. Files
with namespace Name; style ns declarations couldn't be joined together in
this way anyway, and files using namespace Name { ... } style declarations
with use imports outside the block would be invalid on their own,
concatenating them wouldn't break (or fix) this situation. Unless I missed
something?

I must say actually that while I think a use declaration outside a
block-style namespace should be invalid, I'm about 50/50 on whether
something like this should be invalid:

<?php

use \Foo;
namespace Bar;

The current behaviour is certainly wrong, but in this case it might be good
to treat it as if the use declaration was after the ns declaration, rather
than making it invalid (I haven't looked at how easy this would be to
accomplish, though).

johannes

11 years ago by Robert Stoll — view source — reply

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-----Ursprüngliche Nachricht-----
Von: are.you.winning@gmail.com [mailto:are.you.winning@gmail.com] Im Auftrag von Chris Wright
Gesendet: Donnerstag, 13. November 2014 13:13
An: Johannes Schlüter
Cc: Robert Stoll; Adam Harvey; PHP Internals
Betreff: Re: AW: [PHP-DEV] forbid use declaration outside of a namespace in PHP 7

That's still perfectly fine because in your code the use statement
is
not outside of a namespace, it is implicitly in the default namespace.
I am talking about the two following scenarios:

use \Exception;
namespace test;

$e = new Exception(); //would fails because test\Exception is
unknown ->
use was defined outside the namespace

Or
namespace a{
}

use some\UseDeclaration\which\is\outside\of\AnyNamespace as Useless;
//
sorry for begin biased ^^

namespace test{
}

If I get it rigth this might break code by people who, for whatever
reason, combine multiple PHP files into a single one. (cat *.php >
full.php && php full.php)

This would only break if the individual files were already invalid. Files with namespace Name; style ns declarations couldn't
be joined together in this way anyway, and files using namespace Name { ... } style declarations with use imports outside
the block would be invalid on their own, concatenating them wouldn't break (or fix) this situation. Unless I missed
something?

[Robert Stoll]
I share Chris' statement. Combining multiple files would only work if they all use namespace blocks or if they do not use any namespaces at all (and thus are implicitly in the default namespace).

I must say actually that while I think a use declaration outside a block-style namespace should be invalid, I'm about 50/50 on
whether something like this should be invalid:

<?php

use \Foo;
namespace Bar;

The current behaviour is certainly wrong, but in this case it might be good to treat it as if the use declaration was after the
ns declaration, rather than making it invalid (I haven't looked at how easy this would be to accomplish, though).

[Robert Stoll]
I think it would be wrong to start to do such auto-fixes. Syntax which is wrong should emit a fatal error IMO (assuming we would disallow use-statement outside of a namespace). In this case the user has to move the statement down a few lines and it works again. Better than auto-fixing which might introduce unintended bugs.

johannes

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