How would one do s/(.+),(.+)/$1,$0/ using .subst ? -y
On Sun, Apr 19, 2020 at 6:21 PM Tobias Boege <t...@taboege.de> wrote: > On Sun, 19 Apr 2020, yary wrote: > > Question from today's Raku meetup. This works in a way I expect > > > > > 'fosdffgg'.subst(/f+/,"( "~ * ~" )", :g); > > ( f )osd( ff )gg > > > > This one, $0 gets the single f each time > > > > > 'fosdffgg'.subst(/(f+)/,"( $0 )", :g); > > ( f )osd( f )gg > > > > Bug or misunderstanding on my part? > > > > Not a bug. You are calling the subst method on a string. Naturally, in > order > to perform the call, all the arguments you pass have to be evaluated first. > This includes interpolating $0 into the string "( $0 )". You take whatever > $0 contains, put it into a string and pass that into subst. In your case $0 > seemed to confusingly stringify to "f". Consider this instead: > > "oops" ~~ /^(..)/ andthen say 'fosdffgg'.subst(/(f+)/,"( $0 )", :g) > # OUTPUT: «( oo )osd( oo )gg» > > where I deliberately set $0 beforehand. > > Your first variant works because you pass a Callable into subst and when > you > do that subst will repeatedly evaluate the Callable. Not so when you pass a > string value. > > Regards, > Tobias > > -- > "There's an old saying: Don't change anything... ever!" -- Mr. Monk >
