Hi, El sáb., 2 feb. 2019 a las 7:48, ToddAndMargo via perl6-users (< perl6-us...@perl.org>) escribió:
> Hi All, > > How do I use chr inside a regex. In the below, how > do I get rid of $y? > > $ p6 'my Str $x=chr(0x66)~chr(0x77); my Str $y=chr(0x66)~chr(0x77); > $x~~s/ $y /xy/; say $x;' > > If what you want to do is precisely what you are doing, you don't even need to use chr: my $x = "\x66\x77"; $x ~~ s/\x66\x77/xy/; say $x # OUTPUT: «xy» (See the document on quoting: https://docs.perl6.org/language/quoting#Interpolation:_qq) However, if what you want to do is what you _say_ you are doing, my $x = "\x66\x77"; $x ~~ s/$(chr(0x66)~chr(0x77))/xy/; say $x; # OUTPUT: «xy» $() interpolates within a regex, as indicated in the documentation: https://docs.perl6.org/language/regexes#index-entry-regex__Regex_Interpolation-Regex_interpolation Cheers JJ
