in reply to Re: parsing output of UNIX `who` command
in thread parsing output of UNIX `who` command

who|perl -nle '/ /,$x{$`}++}{print"@l"while@l=each%x'

:-)

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Re^3: parsing output of UNIX `who` command (golf)
by cephas (Pilgrim) on Oct 02, 2006 at 13:47 UTC
    who|perl -nle'/ /,$;{$`}++}{print"@l"while@l=each%'
Re^3: parsing output of UNIX `who` command (golf)
by cephas (Pilgrim) on Oct 02, 2006 at 17:50 UTC
    who|perl -ne'/ /,$;{$`}++}{print$_,$/^="*"for%'
      who|perl -nle'/ /,$;{$`.$&}++}{print each%'
      I downloaded this when it was posted and have been poking my brain at it for a little while. Now my brain hurts.
      Any chance you could break it down for us obfunoobs?

        Updated: Added in the forgotten opening brace before the print.

        The parent of your node seems to work, while the first response to it doesn't work the same way. I'll break down the one that works :)

        perldoc perlrun says:

        -n causes Perl to assume the following loop around your program, which makes it iterate over filename argu- ments somewhat like sed -n or awk: LINE: while (<>) { ... # your program goes here }

        So, with the golfed version: who|perl -ne'/ /,$;{$`}++}{print$_,$/^="*"for%', we can take out what was inside the single quotes and put that together manually:

        while(<>){ / /, $;{$`}++}{print $_, $/ ^= "*" for % }

        The interesting thing to notice is now this code isn't properly indented, so let's fix that:

        while(<>){ / /, $;{$`}++ } { print $_, $/ ^= "*" for % }

        What I didn't realize is that apparently -n also apparently adds (?) terminal semicolons (someone correct me if I'm wrong, but with what's happening here certainly seems to be the case), so now it's better re-written as:

        while(<>){ / /, $;{$`}++ } { print $_, $/ ^= "*" for %; }

        Starting to get clearer. Let's just make it crystal clear:

        use English; while( <ARGV> ){ m/\s/; $hash{$PREMATCH}++; } { for( %hash ){ print $_, $INPUT_RECORD_SEPARATOR ^= '*'; } }

        All that's left to explain is that:

        • $PREMATCH contains the username
        • $hash{'username'} counts how many times a user is logged in
        • $INPUT_RECORD_SEPARATOR (by default, a newline) when XOR'd with an asterisk becomes a space, such that the first time through the loop (on the first hash key), "\n" ^ "*" is a space, and now that's stored back in $INPUT_RECORD_SEPARATOR. Next time through the loop (the first hash value), a space XOR'd with an asterisk is a newline, and stored back in $INPUT_RECORD_SEPARATOR.

        Hence the output:

        $username $num_logins $username $num_logins

        Running it through B::Deparse confirms this:

        $ who|perl -MO=Deparse -ne'/ /,$;{$`}++}{print$_,$/^="*"for%' LINE: while (defined($_ = <ARGV>)) { / /, ++$;{$`}; } { foreach $_ (%;) { print $_, $/ ^= '*'; } }


        --chargrill
        s**lil*; $*=join'',sort split q**; s;.*;grr; &&s+(.(.)).+$2$1+; $; = qq-$_-;s,.*,ahc,;$,.=chop for split q,,,reverse;print for($,,$;,$*,$/)