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custom declarator examples in S12 don't work (nested packages) #4639

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p6rt opened this issue Oct 10, 2015 · 2 comments
Open

custom declarator examples in S12 don't work (nested packages) #4639

p6rt opened this issue Oct 10, 2015 · 2 comments
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@p6rt
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@p6rt p6rt commented Oct 10, 2015

Migrated from rt.perl.org#126317 (status was 'new')

Searchable as RT126317$

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@p6rt p6rt commented Oct 10, 2015

From @LLFourn

http://design.perl6.org/S12.html#Custom_Meta-objects

  my module EXPORTHOW​::DECLARE { }
  EXPORTHOW​::DECLARE<controller> = ControllerHOW;

This currently throws "cannot modify an immuteable
EXPORTHOW​::DECLARE". But if we fix that there is another problem​:

my package EXPORTHOW​::DECLARE {
  OUR​::<controller> := Metamodel​::ClassHOW;
};

doesn't work either. The thing that works is​:

my package EXPORTHOW {
  package DECLARE {
  OUR​::<controller> := Metamodel​::ClassHOW;
  }
}

Not sure if the docs are wrong or the implementation though the
following problem seems linked​:

package Foo { package Bar { our sub foo {} }};
package Foo​::Bar { our sub bar {} };
say Foo​::Bar​::<&foo>.package; #-> (Bar)
say Foo​::Bar​::<&bar>.package; #-> (Foo​::Bar)

But if you then try in a new interpreter​:

package Foo { package Bar { our sub foo {} }};
package Foo​::Bar { our sub bar {} };
#!> Redeclaration of routine foo (already defined in package Foo​::Bar)

TL;DR rakudo is not sure which namespace nested packages belong to

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@p6rt p6rt commented Oct 12, 2015

From @LLFourn

I made a pasta error. The final example should be​:

package Foo { package Bar { our sub foo {} }};
package Foo​::Bar { our sub foo {} };
#!> Redeclaration of routine foo (already defined in package Foo​::Bar)

On Sat, Oct 10, 2015 at 11​:41 PM, perl6 via RT <perl6-bugs-followup@​perl.org

wrote​:

Greetings,

This message has been automatically generated in response to the
creation of a trouble ticket regarding​:
"[BUG] custom declarator examples in S12 don't work (nested
packages)",
a summary of which appears below.

There is no need to reply to this message right now. Your ticket has been
assigned an ID of [perl #​126317].

Please include the string​:

     \[perl #&#8203;126317\]

in the subject line of all future correspondence about this issue. To do
so,
you may reply to this message.

                    Thank you,
                    perl6\-bugs\-followup@&#8203;perl\.org

-------------------------------------------------------------------------
http://design.perl6.org/S12.html#Custom_Meta-objects

my module EXPORTHOW&#8203;::DECLARE \{ \}
EXPORTHOW&#8203;::DECLARE\<controller> = ControllerHOW;

This currently throws "cannot modify an immuteable
EXPORTHOW​::DECLARE". But if we fix that there is another problem​:

my package EXPORTHOW​::DECLARE {
OUR​::<controller> := Metamodel​::ClassHOW;
};

doesn't work either. The thing that works is​:

my package EXPORTHOW {
package DECLARE {
OUR​::<controller> := Metamodel​::ClassHOW;
}
}

Not sure if the docs are wrong or the implementation though the
following problem seems linked​:

package Foo { package Bar { our sub foo {} }};
package Foo​::Bar { our sub bar {} };
say Foo​::Bar​::<&foo>.package; #-> (Bar)
say Foo​::Bar​::<&bar>.package; #-> (Foo​::Bar)

But if you then try in a new interpreter​:

package Foo { package Bar { our sub foo {} }};
package Foo​::Bar { our sub bar {} };
#!> Redeclaration of routine foo (already defined in package Foo​::Bar)

TL;DR rakudo is not sure which namespace nested packages belong to

@p6rt p6rt added the Bug label Jan 5, 2020
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