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UNIT::EXPORT and Foo::EXPORT are no longer comparable (Foo::EXPORT === UNIT::EXPORT) #5932
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From @AlexDanielCode: Result (2016.04): Result (HEAD): Bisectable is pointing to rakudo/rakudo@fe2be65 |
From @skidsOn Wed, 28 Dec 2016 19:46:01 -0800, alex.jakimenko@gmail.com wrote:
Since that commit, rakudo/rakudo@4d85cde The bug which 4d85cde9 was trying to address, #128931, pointed out that any two types with the same It is proper that we should have different .WHICH's on different type objects, even if they have $ perl6 -e 'unit module Foo; use nqp; sub g is export { }; nqp::say(nqp::eqaddr(Foo::EXPORT, UNIT::EXPORT))' ...that will either have to be fixed, or worked around. |
The RT System itself - Status changed from 'new' to 'open' |
From @skidsOn Wed, 28 Dec 2016 19:46:01 -0800, alex.jakimenko@gmail.com wrote:
OK, I looked into this some more, and I don't think it is a bug. "unit module" is documented as such: None of that says UNIT and unit module Foo will be the same package, just that The description of the EXPORT and UNIT::EXPORT classes are as such in S11: "Every compunit has a C<UNIT> package, which gets a lexically scoped C<EXPORT> ...note I think there is a typo in S11 where it says: the lexical C<IMPORT> package in C<UNIT>, on the other ...in that IMPORT should be EXPORT. I don't see anything in roast that tests that the two EXPORTs are ===, Now the question is, what was the use case of comparing the two EXPORT Keeping this open so the spec typo may be verified by second opinion and |
Migrated from rt.perl.org#130435 (status was 'open')
Searchable as RT130435$
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