Re: .contains question

On 12/11/23 14:47, Andy Bach wrote:  I have found that when using `say` for debugging, it has been known to print out the previous value of a variable and not the current value.  `print` does not do this. That would certainly be a surprise to me. I'd think I was misunderstanding my program,

Re: .contains question

a bug in say. From: ToddAndMargo via perl6-users Sent: Monday, December 11, 2023 3:24 PM To: perl6-users@perl.org Subject: Re: .contains question CAUTION - EXTERNAL: > "so" will collapse the junction into a Bool. > "say" will append a \n for

Re: .contains question

"so" will collapse the junction into a Bool. "say" will append a \n for you, so you don't have to. On 11 Dec 2023, at 01:52, ToddAndMargo via perl6-users wrote: On 10 Dec 2023, at 21:36, ToddAndMargo via perl6-users wrote: Hi All, my Str $x="abc3defg"; if $x.contains( "a" || "b" ||

Re: .contains question

my $x="abc45def"; my @y=; say so $x.contains(any @y); "so" will collapse the junction into a Bool. "say" will append a \n for you, so you don't have to. > On 11 Dec 2023, at 01:52, ToddAndMargo via perl6-users > wrote: > >>> On 10 Dec 2023, at 21:36, ToddAndMargo via perl6-users >>> wrote:

Re: .contains question

On 10 Dec 2023, at 21:36, ToddAndMargo via perl6-users wrote: Hi All, my Str $x="abc3defg"; if $x.contains( "a" || "b" || "3" ) { print "True\n"; } else { print "False\n" }; True Is there a way to tell .contains that you want to know if any of a sequence characters is in a string other

Re: .contains question

my @letters = ; if $x.contains(any @letters) { ... > On 10 Dec 2023, at 21:36, ToddAndMargo via perl6-users > wrote: > > Hi All, > > my Str $x="abc3defg"; if $x.contains( "a" || "b" || "3" ) { print "True\n"; > } else { print "False\n" }; > True > > Is there a way to tell .contains that

.contains question

Hi All, my Str $x="abc3defg"; if $x.contains( "a" || "b" || "3" ) { print "True\n"; } else { print "False\n" }; True Is there a way to tell .contains that you want to know if any of a sequence characters is in a string other that repeating || over and over. Any [a..z] or [0..9] option?

Re: contains question

Yup, you can use it as a method call like this: perl6 -e '"foobar".&[~~](/foo/).say' ｢foo｣ HTH - Timo

Re: contains question

On Mon, Jun 12, 2017 at 4:07 PM, Elizabeth Mattijsen wrote: >> On 12 Jun 2017, at 22:04, Will Coleda wrote: >> On Mon, Jun 12, 2017 at 5:17 AM, Francesco Rivetti wrote: >>> if you can: >>> >>> $s ~~ "foo" >>> $s ~~ /foo/ >>> >>> then wouldn't be

Re: contains question

On Mon, Jun 12, 2017 at 5:17 AM, Francesco Rivetti wrote: > if you can: > > $s ~~ "foo" > $s ~~ /foo/ > > then wouldn't be good to have also: > > $s.contains("foo"); > $s.contains(/foo/); The latter is currently available as: > "foobar".match(/'foo'/); ｢foo｣ > IOW, overload

Re: contains question

On 06/12/2017 01:42 AM, Elizabeth Mattijsen wrote: On 12 Jun 2017, at 01:27, ToddAndMargo wrote: perl6 -e 'my $x = "\t"; if $x !~~ /<[A..Z a..z 0..9]>/ {say "out"} else {say "in"}' Would this be easier to do with $x.contains? Or would it be too worky? .contains

Re: contains question

Thinking about this... > On 12 Jun 2017, at 11:17, Francesco Rivetti wrote: > > if you can: > > $s ~~ "foo" > $s ~~ /foo/ > > then wouldn't be good to have also: > > $s.contains("foo"); > $s.contains(/foo/); > > IOW, overload .contains() with Str and Regex > > F > > On

Re: contains question

if you can: $s ~~ "foo" $s ~~ /foo/ then wouldn't be good to have also: $s.contains("foo"); $s.contains(/foo/); IOW, overload .contains() with Str and Regex F On 06/12/2017 10:42 AM, Elizabeth Mattijsen wrote: On 12 Jun 2017, at 01:27, ToddAndMargo wrote: perl6 -e

Re: contains question

> On 12 Jun 2017, at 01:27, ToddAndMargo wrote: > perl6 -e 'my $x = "\t"; if $x !~~ /<[A..Z a..z 0..9]>/ {say "out"} else {say > "in"}' > > Would this be easier to do with $x.contains? Or would it > be too worky? .contains only takes a *single string* to look up. So

contains question

Hi All, perl6 -e 'my $x = "\t"; if $x !~~ /<[A..Z a..z 0..9]>/ {say "out"} else {say "in"}' Would this be easier to do with $x.contains? Or would it be too worky? Many thanks, -T