Date: Wed, 28 Jul 2021 11:36:33 -0600 From: Warner Losh <imp@bsdimp.com> To: Michael Butler <imb@protected-networks.net> Cc: freebsd-current <freebsd-current@freebsd.org> Subject: Re: awk behaviour? Message-ID: <CANCZdfp1nTx0%2B4ZKWYOPN9WtBzDPHYRnzLL58=Ni7giwFQFmfw@mail.gmail.com> In-Reply-To: <8e1a8e3c-b062-7749-ceab-e500c1ab758e@protected-networks.net> References: <8e1a8e3c-b062-7749-ceab-e500c1ab758e@protected-networks.net>
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On Wed, Jul 28, 2021 at 11:31 AM Michael Butler via freebsd-current <
freebsd-current@freebsd.org> wrote:
> I tripped over this while trying to build a local release ..
>
> imb@toshi:/home/imb> pkg --version | awk -F. '{print $$1 * 10000 + $$2 *
> 100 + $$3}'
> 10001
>
> imb@toshi:/home/imb> pkg --version
> 1.17.1
>
> Is this expected?
>
Why $$ instead of $? $ isn't expanded in '' expressions, so doubling isn't
necessary
unlike in make... With single quotes it works for me:
% pkg --version | awk -F. '{print $1 * 10000 + $2 * 100 + $3}'
11603
% pkg --version
1.16.3
In awk $$n is $($n), so $$ in this context would evaluate $1 to 1 and then
$1 to be 1. And then $2 to be 16
and then $17 to be 0 and then $3 to be 1 and then $1 to be 1 which leads to
10001.
Warner
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