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isa method errors when called on a subset type object #6510

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p6rt opened this issue Sep 12, 2017 · 7 comments
Closed

isa method errors when called on a subset type object #6510

p6rt opened this issue Sep 12, 2017 · 7 comments
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@p6rt
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p6rt commented Sep 12, 2017

Migrated from rt.perl.org#132073 (status was 'resolved')

Searchable as RT132073$

@p6rt
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p6rt commented Sep 12, 2017

From @dwarring

current behavior of isa method on a subset​:

% perl6 -v
This is Rakudo version 2017.08-110-g5f3350656 built on MoarVM version
2017.08.1-156-g4
9b90b99
implementing Perl 6.c.
% perl6 -e'subset S of Int; say S.isa(Int)'
Cannot resolve caller isa(Perl6​::Metamodel​::SubsetHOW​: S, Int); none of
these signatur
es match​:
  (Mu \SELF​: Mu $type, *%_)
  (Mu \SELF​: Str​:D $name, *%_)
in block <unit> at -e line 1

The documentation https://docs.perl6.org/routine/isa implies this should
return True.

@p6rt
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p6rt commented Sep 16, 2017

From @AlexDaniel

What about this commit? rakudo/rakudo@0704cd9

On 2017-09-12 13​:55​:51, david.warring wrote​:

current behavior of isa method on a subset​:

% perl6 -v
This is Rakudo version 2017.08-110-g5f3350656 built on MoarVM version
2017.08.1-156-g4
9b90b99
implementing Perl 6.c.
% perl6 -e'subset S of Int; say S.isa(Int)'
Cannot resolve caller isa(Perl6​::Metamodel​::SubsetHOW​: S, Int); none of
these signatur
es match​:
(Mu \SELF​: Mu $type, *%_)
(Mu \SELF​: Str​:D $name, *%_)
in block <unit> at -e line 1

The documentation https://docs.perl6.org/routine/isa implies this should
return True.

@p6rt
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p6rt commented Sep 16, 2017

The RT System itself - Status changed from 'new' to 'open'

@p6rt
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p6rt commented Sep 16, 2017

From @dwarring

After that commit​: subset S of Int; S.isa(True) returns true as expected.

I've noticed a quibble with subset of a subset​:

perl6 -e'subset S of Int; subset S2 of S; say S2.isa(S)'
False

Should be True.

On Sun, Sep 17, 2017 at 3​:17 AM, Aleks-Daniel Jakimenko-Aleksejev via RT <
perl6-bugs-followup@​perl.org> wrote​:

What about this commit?
rakudo/rakudo@0704cd9
ef1c5fe711

On 2017-09-12 13​:55​:51, david.warring wrote​:

current behavior of isa method on a subset​:

% perl6 -v
This is Rakudo version 2017.08-110-g5f3350656 built on MoarVM version
2017.08.1-156-g4
9b90b99
implementing Perl 6.c.
% perl6 -e'subset S of Int; say S.isa(Int)'
Cannot resolve caller isa(Perl6​::Metamodel​::SubsetHOW​: S, Int); none of
these signatur
es match​:
(Mu \SELF​: Mu $type, *%_)
(Mu \SELF​: Str​:D $name, *%_)
in block <unit> at -e line 1

The documentation https://docs.perl6.org/routine/isa implies this should
return True.

@p6rt
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p6rt commented Sep 16, 2017

From @dwarring

Tests added with roast commit
Raku/roast@d776a06

One remaining todo test for the subset of a subset case, ie​:

subset S of Int; subset S2 of S; say S2.isa(S)

On Sun, Sep 17, 2017 at 6​:34 AM, David Warring <david.warring@​gmail.com>
wrote​:

After that commit​: subset S of Int; S.isa(True) returns true as expected.

I've noticed a quibble with subset of a subset​:

perl6 -e'subset S of Int; subset S2 of S; say S2.isa(S)'
False

Should be True.

On Sun, Sep 17, 2017 at 3​:17 AM, Aleks-Daniel Jakimenko-Aleksejev via RT <
perl6-bugs-followup@​perl.org> wrote​:

What about this commit?
rakudo/rakudo@0704cd97226e63001943
426666c88cef1c5fe711

On 2017-09-12 13​:55​:51, david.warring wrote​:

current behavior of isa method on a subset​:

% perl6 -v
This is Rakudo version 2017.08-110-g5f3350656 built on MoarVM version
2017.08.1-156-g4
9b90b99
implementing Perl 6.c.
% perl6 -e'subset S of Int; say S.isa(Int)'
Cannot resolve caller isa(Perl6​::Metamodel​::SubsetHOW​: S, Int); none of
these signatur
es match​:
(Mu \SELF​: Mu $type, *%_)
(Mu \SELF​: Str​:D $name, *%_)
in block <unit> at -e line 1

The documentation https://docs.perl6.org/routine/isa implies this
should
return True.

@p6rt
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p6rt commented Sep 20, 2017

From @dwarring

Final issues resolved with rakudo/rakudo@cee1be2

On Sat, 16 Sep 2017 12​:12​:35 -0700, david.warring wrote​:

Tests added with roast commit
Raku/roast@d776a06

One remaining todo test for the subset of a subset case, ie​:

subset S of Int; subset S2 of S; say S2.isa(S)

On Sun, Sep 17, 2017 at 6​:34 AM, David Warring
<david.warring@​gmail.com>
wrote​:

After that commit​: subset S of Int; S.isa(True) returns true as
expected.

I've noticed a quibble with subset of a subset​:

perl6 -e'subset S of Int; subset S2 of S; say S2.isa(S)'
False

Should be True.

On Sun, Sep 17, 2017 at 3​:17 AM, Aleks-Daniel Jakimenko-Aleksejev via
RT <
perl6-bugs-followup@​perl.org> wrote​:

What about this commit?
rakudo/rakudo@0704cd97226e63001943
426666c88cef1c5fe711

On 2017-09-12 13​:55​:51, david.warring wrote​:

current behavior of isa method on a subset​:

% perl6 -v
This is Rakudo version 2017.08-110-g5f3350656 built on MoarVM
version
2017.08.1-156-g4
9b90b99
implementing Perl 6.c.
% perl6 -e'subset S of Int; say S.isa(Int)'
Cannot resolve caller isa(Perl6​::Metamodel​::SubsetHOW​: S, Int);
none of
these signatur
es match​:
(Mu \SELF​: Mu $type, *%_)
(Mu \SELF​: Str​:D $name, *%_)
in block <unit> at -e line 1

The documentation https://docs.perl6.org/routine/isa implies this
should
return True.

@p6rt
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p6rt commented Sep 20, 2017

@dwarring - Status changed from 'open' to 'resolved'

@p6rt p6rt closed this as completed Sep 20, 2017
@p6rt p6rt added the Bug label Jan 5, 2020
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