Hello,
I have a few questions about the regexes. I'm new to perl 6
and I don't know perl 5. I'm proficient with grep/sed.
I installed Rakudo with the defaults on Linux (Mandriva 2010.1)
and Windows 7. I get identical output for all these questions
with the two latest Rakudo versions on Linux (2014.01 2013.12)
and Windows (2014.01 2013.09).
Questions
(1) :global
This works as expected:
say ~( 'aaa'.match(/a/,:g) ) # a a a
but this doesn't work:
say ?( 'aaa' ~~ m:g/a/) # False
Is the second form m:g/.../ implemented?
if it is, can you give me an example
(2) capture to lexical variables, external aliasing
>From the synopsis:
You may capture to existing lexical variables;
such variables may already be visible from an
outer scope, or may be declared within the
regex via a :my declaration.
my $x; / $x = [...] / # capture to outer lexical $x
/ :my $x; $x = [...] / # capture to our own lexical $x
Here I have two examples and they don't work, why?
is this feature implemented?
my $x; 'a' ~~ / $x=. /; say $x; # (Any)
'a' ~~ / :my $x; $x = . {say $x} / # [no output]
(3) arrays
None of these four cases work and I don't know why:
my @x; 'abc' ~~ / @x = .+ /; say @x[0]; # (Any)
'abc' ~~ /^ @<x>=.+ $/; say @<x>; # Nil
my @x; 'abc' ~~ /@x = (.)+/; # Could not find sub cuid_1_1384375491.61906
'abc' ~~ / @<x> = (.)+ /; # Could not find sub cuid_1_1391263931.391
(4) submatch
I'm somewhat puzzled what a submatch is.
This works:
'ab' ~~ / (..) { say ~($0 ~~ /b/) } / # b
But can a submatch be immediately nested in a regex like this:
'ab' ~~ /(..) [$0 ~~ b] }/ # doesn't work
(5) interpolation
how do I get a variable to interpolate into <[]> ?
doesn't work:
my $x='a';
say ~('aaa' ~~ /^ <[$x]>+ $/) # [no output]
(6) why does this enter an infinite loop instead of matching?
'a' ~~ /^ [.*? {say 'ok'}]+ $/
(7) why don't these produce the same output?
say 'ab'.subst(/ '' <?before b> /,'c'); # ab
say 'ab'.subst(/ <?> <?before b> /,'c'); # acb
say 'ab'.subst(/ <?before b> /,'c'); # acb
Similar question for these:
'abbbb' ~~ / (b ** 0..3)/; say ~$0; # [no output]
'abbbb' ~~ / (b ** 1..3)/; say ~$0; # bbb
(8) implicit scanning at the beginning
(a) regex/token/rule
I don't understand the behavior of regex/token/rule in
these cases, for I thought they didn't implicit scan:
'a11' ~~ regex { (\d+) }; say "$0"; # 11
'a11' ~~ token { (\d+) }; say "$0"; # 11
'a11' ~~ rule { (\d+) }; say "$0"; # 11
the synopsis seems to suggest the opposite:
...
$string ~~ regex { \d+ }
$string ~~ token { \d+ }
$string ~~ rule { \d+ }
and these are equivalent to
$string ~~ m/^ \d+ $/;
$string ~~ m/^ \d+: $/;
$string ~~ m/^ <.ws> \d+: <.ws> $/;
(b) I find this behavior unexpected:
say ?('ab' ~~ / <?before a> /); # True
say ?('ab' ~~ / <?before b> /); # True
say ?('ab' ~~ / <?before a> <?before b> /); # False
In each case there is scanning at the very beginning, but
in the 3rd case there is no scanning before <?before b> as
if <?before a> had the side effect of cancelling it
in spite of its "?"
i.e I was expecting "?" to make <?before a> side effect free
(9) Please verify if the following section of the
"Regexes and Rules" synopsis contains an error:
"In contrast, if the outer quantified structure is a
capturing structure (i.e. a subpattern) then..."
my $text = "foo:food fool\nbar:bard barb";
# $0-----------------------
# | |
# | $0[0] $0[1]--- |
# | | | | | |
$text ~~ m/ ( (\w+) \: (\w+ \h*)* \n ) ** 2..* /;
The error seems that $text lacks a final \n. It works as expected
if you add it. You should also verify the diagram above, it may,
or not, contain an error. I just tried it thus instead and it seems to work:
my $text = "foo:food fool\nbar:bard barb\n";
# $0-------------------------------
# | |
# |($0[0][0],$0[0][1]) 1st iter. |
# |($0[1][0],$0[1][1]) 2nd iter. |
# || | | | |
$text ~~ m/((\w+) \: (\w+ \h*)* \n) ** 2..*/;
say "$0[0][0]"; # foo
say "$0[0][1]"; # food fool
say "$0[1][0]"; # bar
say "$0[1][1]"; # bard barb
thank you
