On Sat Apr 16 05:37:53 2011, masak wrote:
> <masak> awww.
> <masak> who would have guessed, declaring a sub infix:<;> causes
problems...
> <masak> rakudo: our sub infix:<;>($lhs, $rhs) {}; { say "A"; say "B" }
> <p6eval> rakudo 5ac05e: ( no output )
> <masak> :(
> <masak> rakudo: our sub infix:<;>($lhs, $rhs) { say $lhs, $rhs }; 1;
2; 3
> <p6eval> rakudo 5ac05e: OUTPUT«infix:<;>1Bool::True2Bool::True3»
> <masak> I believe this is backwards. the stopper ';' should take
> precedence over any declared infixes, unless we're inside a
> parenthetical thingy.
> * masak submits rakudobug
<jnthn> What does STD think?
<jnthn> std: sub infix:<;>($, $) {}; { say "A"; say "B" }
<p6eval> std 4608239: OUTPUT«ok 00:01 122m»
<jnthn> I'm thinking that example doesn't force it to reveal what it's
parsing that ; as though
<jnthn> And wondering how to construct one that does :)
<colomon> without executing the code, I can't see how to do it.
<jnthn> oh, maybe you can do it with statement modifiers
<jnthn> std: sub infix:<;>($, $) {}; { say "A" if 1; say "B" if 1; }
<p6eval> std 4608239: OUTPUT«ok 00:01 122m»
<jnthn> I expect that'd fail if the ; wasn't being parsed as a statement
sep there.
<masak> jnthn++
* masak submits rakudobug
<jnthn> You already did submit the Rakudo bug! :P
<masak> apparently I'm getting senile.
<masak> ok, I'll just amend that ticket, then.
<colomon> std: (say "A" if 1) + 10
<p6eval> std 4608239: OUTPUT«ok 00:01 121m»
<colomon> jnthn: are you sure?
<masak> colomon: yes, it's the 'if' twice that makes it work.
<jnthn> std: say "A" if 1 + say "B" if 1
<p6eval> std 4608239: OUTPUT«[31m===[0mSORRY![31m===[0mConfused [...]
FAILED 00:01 120m»
