"Match objects are the result of a successful regex match, this does
include any zero-width match. They store a reference to the original
string (.orig), positional and named captures, the positions of the
start and end of the match in the original string, and a payload
referred to as AST (abstract syntax tree), which can be used to build
data structures from complex regexes and grammars." -
https://docs.perl6.org/type/Match

When you say $0, you're using the Match. If you want the Str version,
explicitly cast it to Str with:

~$0

or

$0.Str

Regards.

On Thu, Dec 20, 2018 at 5:17 PM ToddAndMargo via perl6-users
<perl6-us...@perl.org> wrote:
>
> >> El jue., 20 dic. 2018 21:43, ToddAndMargo via perl6-users
> >> <perl6-us...@perl.org <mailto:perl6-us...@perl.org>> escribió:
> >>
> >>     Hi All,
> >>
> >>     Exactly what is type "Match"?
> >>
> >>     Here I want $D0..$D3 to only be strings.  And it throws a match error.
> >>
> >>     $ p6 'my $x="11.2.3.4"; my Str $D0; my Str $D1; my Str $D2; my Str $D3;
> >>     $x~~m{ (<:N>) [.] (\d+) [.] (\d+) [.] (\d+) }; $D0 = $0; $D1 = $1;
> >>     $D2 =
> >>     $2; $D3 = $3; print "$D0 $D1 $D2 $D3\n";'
> >>
> >>     Type check failed in assignment to $D0; expected Str but got Match
> >>     (Match.new(from => 1, made ...)
> >>         in block <unit> at -e line 1
> >>
> >>     Here is my work around:
> >>
> >>     $ p6 'my $x="11.2.3.4"; my Str $D0; my Str $D1; my Str $D2; my Str $D3;
> >>     $x~~m{ (<:N>+) [.] (\d+) [.] (\d+) [.] (\d+) }; $D0 = $0.Str; $D1 =
> >>     $1.Str; $D2 = $2.Str; $D3 = $3.Str; print "$D0 $D1 $D2 $D3\n";'
> >>     11 2 3 4
> >>
> >>
> >>     Many thanks,
> >>     -T
>
> On 12/20/18 2:08 PM, JJ Merelo wrote:
> > Put a wriggly ~ in front of $0 to turn it into a Str; it's the Str
> > contextualizer
> >
>
> Hi JJ,
>
> You did not actually answer the question I asked.  What is type "Match"?
>
> And I am missing something in your answer
>
> This works:
>
> $ p6 'my $x="11.2."; my Str $D0; my Str $D1; $x~~m{ (<:N>+) [.] (\d+) };
> $D0 = $0.Str; $D1 = $1.Str;  print "$D0 $D1\n";'
> 11 2
>
>
> This does not.  One with a space after the ~, one without it.
>
> $ p6 'my $x="11.2."; my Str $D0; my Str $D1; $x~~m{ (<:N>+) [.] (\d+) };
> $D0 ~$0; $D1 ~ $1;  print "$D0 $D1\n";'
> WARNINGS for -e:
> Useless use of "~" in expression "$D1 ~ $1" in sink context (line 1)
> Useless use of "~" in expression "$D0 ~$0" in sink context (line 1)
> Use of uninitialized value of type Str in string context.
> Methods .^name, .perl, .gist, or .say can be used to stringify it to
> something meaningful.
>    in block <unit> at -e line 1
> Use of uninitialized value of type Str in string context.
> Methods .^name, .perl, .gist, or .say can be used to stringify it to
> something meaningful.
>    in block <unit> at -e line 1
> Use of uninitialized value of type Str in string context.
> Methods .^name, .perl, .gist, or .say can be used to stringify it to
> something meaningful.
>    in block <unit> at -e line 1
> Use of uninitialized value of type Str in string context.
> Methods .^name, .perl, .gist, or .say can be used to stringify it to
> something meaningful.
>    in block <unit> at -e line 1
>
> I am confused,
> -T

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