Calling subroutine with a scalar

Rodster001 has asked for the wisdom of the Perl Monks concerning the following question:

Hi Monks, I know this has been asked endlessly, but I am doing it slightly different than I have done before and I cannot quite get the syntax right.

my $res = new Some::Package({ foo => bar })->subroutine;
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What I want to do is call "subroutine" by variable. For example:


my $sub = "subroutine";
my $res = new Some::Package({ foo => bar })->( $sub );
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That doesn't work, I've tried several variations of that (along with \&) and I can't quite get it. Can someone enlighten me?

Thanks!

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Re: Calling subroutine with a scalar by philipbailey (Curate) on May 06, 2015 at 18:38 UTC
Try `my $res = new Some::Package({ foo => bar })->$sub;` Or better: `my $res = Some::Package->new({ foo => bar })->$sub;`	[reply] [d/l] [select]
Re^2: Calling subroutine with a scalar by Rodster001 (Pilgrim) on May 06, 2015 at 18:53 UTC
That did it. Thank you. Why is it better to call new that way?	[reply]
Re^3: Calling subroutine with a scalar by jeffa (Bishop) on May 06, 2015 at 18:58 UTC
You can learn why here: http://modernperlbooks.com/mt/2009/08/the-problems-with-indirect-object-notation.html jeffa L-LL-L--L-LL-L--L-LL-L-- -R--R-RR-R--R-RR-R--R-RR B--B--B--B--B--B--B--B-- H---H---H---H---H---H--- (the triplet paradiddle with high-hat)	[reply] [d/l]