From @cognominal

A combination of many conditions seem necessary for this to happen :
the lexical must braced within a string in a return statement of a
recursive function.

$ perl6 -e 'sub t($p) {   t $p-1  if $p-1 > 0;  return "{$p}" }; say t
3'1$ perl6 -e 'sub t($p) {   t $p-1  if $p-1 > 0;  return "$p" };
say t 3'3

--
cognominal stef

From @jnthn

On Fri Nov 25 07​:01​:02 2011, cognominal wrote​:

A combination of many conditions seem necessary for this to happen :
the lexical must braced within a string in a return statement of a
recursive function.

$ perl6 -e 'sub t($p) { � t $p-1 �if $p-1 > 0; �return "{$p}" }; say t
3'1$ perl6 -e 'sub t($p) { � t $p-1 �if $p-1 > 0; �return "$p" };
say t 3'3

Fixed​:

sub t($p) { t $p-1 if $p-1 > 0; return "{$p}" }; say t 3
3
sub t($p) { t $p-1 if $p-1 > 0; return "$p" }; say t 3
3

Needs tests.

/jnthn

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

From @moritz

Now tested in S02-literals/misc-interpolation.t

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