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-follo...@perl.org> wrote: > What about this commit? > https://github.com/rakudo/rakudo/commit/0704cd97226e63001943426666c88c > 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. > >
