On 08/14/2018 06:33 PM, yary wrote:
"so" coerces to True or False. "say /c/" would output the match object
"say so /c/" says True. Depends on what you want to see.
" $x ?? $y !! $z" is a shortcut - "if $x use value of $y else use value
of $z" and ought to be used for the final value.
You may
"so" coerces to True or False. "say /c/" would output the match object "say
so /c/" says True. Depends on what you want to see.
" $x ?? $y !! $z" is a shortcut - "if $x use value of $y else use value of
$z" and ought to be used for the final value.
You may know it in perl5 as "$result = $x ? $y :
On 08/14/2018 08:29 AM, yary wrote:
Or, store the string in $_, and take advantage of less to type-
perl6 -e '$_="abc"; say so /z/; say so /b/; s/c/defg/ ?? .say !! say
"Failed!"'
-y
Thank you!
Well I can see it working, but what does
"so"
"??"
".say"
"!!"
do?
My actual
Or, store the string in $_, and take advantage of less to type-
perl6 -e '$_="abc"; say so /z/; say so /b/; s/c/defg/ ?? .say !! say
"Failed!"'
-y
On Tue, Aug 14, 2018 at 4:17 AM, ToddAndMargo wrote:
> > On 14/08/18 13:08, ToddAndMargo wrote:
> >> Hi All,
> >>
> >> The Perl 5 guys have it
> On 14/08/18 13:08, ToddAndMargo wrote:
>> Hi All,
>>
>> The Perl 5 guys have it pounded into my head that I
>> always had to check my substitutions to see if they
>> worked if not working would crash the program.
>>
>> So in Perl 6 I have:
>>
>> $ p6 'my $x="abc"; if s/b/z/ {say "sub
You're putting your starting string in a variable, $x, but aren't
telling the s/// operator specifically what to operate on, so it
defaults to $_, which is still at its default value.
On 14/08/18 13:08, ToddAndMargo wrote:
> Hi All,
>
> The Perl 5 guys have it pounded into my head that I
>
