#          1         2         3
# 1234567890123456789012345678901234567
  while(<STDIN>){print if m/$ARGV[0]/;}
  /$ARGV[0]/&&print while<STDIN>
# 123456789012345678901234567890
#          1         2         3
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($q=q:Sq=~/;[c](.)(.)/;chr(-||-|5+lengthSq)`"S|oS2"`map{chr |+ord
}map{substrSq`S_+|`|}3E|-|`7**2-3:)=~y+S|`+$1,++print+eval$q,q,a,
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Why?

Because I'm trying to learn

Also, thanks, I was not aware of the && operator

#   5    0    5    0    5 7
$p=shift;print grep/$p/,<>;
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and works for any number of files as well as STDIN

However, the content of all the files will be "slurped" into memory because <> is in list context.

To read line by line:

#   5    0    5    0    5    0 2
$p=shift;while(<>){print if/$p/}
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I like this answer because It still allows the default filenames on commandline arguments to stdin behavior with <>.

So maybe you want to have a look at -l -a -n -e -p ...(and so on) options in perlrun ?

e.g.

-n

causes Perl to assume the following loop around your program, which makes it iterate over filename arguments somewhat like sed -n or awk:

      LINE:
        while (<>) {
        ...        # your program goes here
        }
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Note that the lines are not printed by default. See -p to have lines printed. If a file named by an argument cannot be opened for some reason, Perl warns you about it and moves on to the next file.

update

As replacement for

grep PATTERN *

try

perl -ne'/PATTERN/&&print' *

for a start. :)

23:

#   5    0    5    0  3
print grep/$ARGV[0]/,<>;
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This dosen't work for me because perl tries to interpret the pattern as a file name, but I found it does not use this behavior when using <STDIN> instead of <>.

29:

#   5    0    5    0    5   9
print grep/$ARGV[0]/,<STDIN>;
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I cheat a little and assume only one argument on the command line:

print grep/@ARGV/,<STDIN>;
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Nevermind... it took a sec to figure out I had to pass an arg in ;)

#   5    0   4 
print while<>;
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