Shawn H Corey writes:
> On Fri, 25 Apr 2014 12:39:21 -0700
> John SJ Anderson wrote:
>
>> Perl doesn't charge you by the lines of code you use, so doing this:
>>
>>my $re = shift;
>>$re = qr/$re/;
>>
>> is just fine.
>
> This also works:
>
> my $re = qr/$ARGV[0]/;
> shift @ARGV
Uri Guttman writes:
> On 04/25/2014 12:55 PM, Harry Putnam wrote:
>> Uri Guttman writes:
>>
>>> why would you expect anything but 1 as the value? shift will return 1
>>> value or undef. that anon array will be dereferenced to an array with
>>> 1 entry. the array is in scalar context which return
*puts on List Mom hat*
Okay, we're done here. The original question was asked and answered (a
couple times). Let's move on and get back to dealing with Perl
questions, not personalities.
thanks,
john.
*takes List Mom hat back off*
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For a
On 26/04/2014 02:16, Uri Guttman wrote:
On 04/25/2014 12:55 PM, Harry Putnam wrote:
Uri Guttman writes:
why would you expect anything but 1 as the value? shift will return 1
value or undef. that anon array will be dereferenced to an array with
1 entry. the array is in scalar context which ret
On 04/25/2014 12:55 PM, Harry Putnam wrote:
Uri Guttman writes:
why would you expect anything but 1 as the value? shift will return 1
value or undef. that anon array will be dereferenced to an array with
1 entry. the array is in scalar context which returns its
size. elementary!
On Fri, 25 Apr 2014 12:39:21 -0700
John SJ Anderson wrote:
> Perl doesn't charge you by the lines of code you use, so doing this:
>
>my $re = shift;
>$re = qr/$re/;
>
> is just fine.
This also works:
my $re = qr/$ARGV[0]/;
shift @ARGV;
--
Don't stop where the ink does.
On Fri, Apr 25, 2014 at 12:18 PM, Harry Putnam wrote:
> About just using 'shift': How I happened to be using that weird
> looking notation at all was because I thought it was a way to avoid
> the extra step of compiling the rgx in qr// after shifting.
My advice to you on this point would be tha
John SJ Anderson writes:
> In general, anywhere you're doing '@{[shift]}', unless you REALLY know
> what's going on and why you'd want to do that ... instead just do
> 'shift'.
Thanks for the in depth answer. Very helpful. You've helped fill
some tragic holes in my basic knowledge of perl. I'
Uri Guttman writes:
> why would you expect anything but 1 as the value? shift will return 1
> value or undef. that anon array will be dereferenced to an array with
> 1 entry. the array is in scalar context which returns its
> size. elementary!
^ my dear Watson.
Thank you Mr U
On 04/24/2014 11:40 PM, Harry Putnam wrote:
Some simple code that is similar to other code I've written and used
is returning something I don't understand.
I've used the notation below for shifting off elements of AR many times
but don't recall seeing this output. I think I know what is happeni
On Thu, Apr 24, 2014 at 8:40 PM, Harry Putnam wrote:
> my $dir2sr = @{[shift]};
This shifts an element off @ARGV, makes an array reference containing
a single element (the value that was just shifted off), then
deferences that arrayref to create an array (which still contains a
single element, t
Some simple code that is similar to other code I've written and used
is returning something I don't understand.
I've used the notation below for shifting off elements of AR many times
but don't recall seeing this output. I think I know what is happening
but I don't understand why.
Is the second
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