in reply to Answer: How do I remove whitespace at the beginning or end of my string?
in thread regular expressions

If you want to preserve internal whitespace, try
s/ #replace ^[\s]* #the start of the string followed by any number of spaces (.*) #zero or more characters of anything, stuff into $1. This +is the actual string we want. (?<!\s) #look-behind assertion to get to the last non-whitespace ch +ar \s*$ #match whitespace until EOL /$1/x; #replace with the (.*)

Update: If you value your CPU time, use one of the other regular expressions instead. I got caught up in using the .*, leading to the need for the look-behind assertion. Better answers can be found here. For the interested,

Benchmark: timing 100000 iterations of my_long_one, one_liner, two_lin +er... my_long_one: 6 wallclock secs ( 4.62 usr + 0.00 sys = 4.62 CPU) @ 2 +1659.09/s (n=100000) one_liner: 3 wallclock secs ( 3.89 usr + 0.00 sys = 3.89 CPU) @ 256 +73.94/s (n=100000) two_liner: 1 wallclock secs ( 2.16 usr + 0.00 sys = 2.16 CPU) @ 462 +10.72/s (n=100000) Using my_long_one => s/^[\s]*(.*)(?<!\s)\s*$/$1/ one_liner => s/^\s+|\s+$//g two_liner => s/^\s+//g; s/s+$//g