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
On 12/20/18 2:34 PM, Will Coleda wrote:
"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.
So similar to type "Mu". Kind of all things but doesn't take on
a particular type until you assign it to something.
I will stick with $0.Str as ~$0 does not work with "perl6 -e".
And .Str means something to me immediately. ~ means concatenate
strings to me, so it takes a bit of thinking.
Thank you!
