Interpolate a variable into a regular expression

Question

I am used to Perl but a Perl 6 newbie

I want to host a regular expression in a text variable, like I would have done in perl5:

my $a = 'abababa';
my $b = '^aba';
if ($a =~ m/$b/) {
        print "True\n";
} else {
        print "False\n";
}

But if I do the same in Perl6 it doesn't work:

my $a = 'abababa';
my $b = '^aba';

say so $a ~~ /^aba/;  # True
say so $a ~~ /$b/;    # False

I'm puzzled... What am I missing?

Answer 1

You need to have a closer look at Quoting Constructs.

For this case, enclose the part of the LHS that is a separate token with angle brackets or <{ and }>:

my $a = 'abababa';
my $b = '^aba';
say so $a ~~ /<$b>/;       # True, starts with aba
say so $a ~~ /<{$b}>/;     # True, starts with aba

my $c = '<[0..5]>'
say so $a ~~ /<$c>/;       # False, no digits 1 to 5 in $a
say so $a ~~ /<{$c}>/;     # False, no digits 1 to 5 in $a

Another story is when you need to pass a variable into a limiting quantifier. That is where you need to only use braces:

my $ok = "12345678";
my $not_ok = "1234567";
my $min = 8;
say so $ok ~~ / ^ \d ** {$min .. *} $ /;         # True, the string consists of 8 or more digits
say so $not_ok ~~ / ^ \d ** {$min .. *} $ /;     # False, there are 7 digits only

Answer 2

Is there a reason why you don't pick the regex object for these types of uses?

my $a = 'abababa';
my $b = rx/^aba/;

say so $a ~~ /^aba/;  # True
say so $a ~~ $b;     # True