can you explain this perl oneliner

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

Hi, This perl one liner will print top ten memory using process. can you explain how this perl oneliner works?

svmon -Pt15 | perl -e 'while(<>){print if($.==2||$&&&!$s++);
$.=0 if(/^-+$/)}'
[download]

thanks

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Re: can you explain this perl oneliner by Utilitarian (Vicar) on Jan 23, 2010 at 02:24 UTC
OK, so you're using AIX, not everyone is familiar with it, so consider providing sample input to the Perl script `svmon -P, mem usage for the following PIDs -t$n, Displays memory usage statistics for the top $n processes` [download] So that's our input. We skip the first line by insisting that $. (input line number) is 2 Then allow following lines with the OR `$&` (last pattern match, from line 2 on exists) There is also a redundant check on `!$s++` which only can return true when `$s=0` however if the line consists of dashes "-" reset the counter to zero. For more precise answers try adding the input, the result of `svmon -Pt15` to the enquiry. Hope this helps. UPDATEBy the way this code is horribly inefficient and seems to be written for obscurity more than for effectiveness `print "Good ",qw(night morning afternoon evening)[(localtime)[2]/6]," fellow monks."`	[reply] [d/l] [select]
Re: can you explain this perl oneliner by Anonymous Monk on Jan 23, 2010 at 02:16 UTC
`$ perl -MO=Deparse -e "while(<>){print if($.==2\|\|$&&&!$s++);$.=0 if(/^ +-+$/)}" while (defined($_ = <ARGV>)) { print $_ if $. == 2 or $& and not $s++; $. = 0 if /^-+$/; } -e syntax OK` [download] It prints the 2rd line read, and 2nd line after every line of ----. Also, it prints the first line after the first line of ----.	[reply] [d/l]