chop vs chomp

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

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Re: chop vs chomp by pg (Canon) on Oct 30, 2005 at 23:23 UTC
Try this demo: `use strict; use warnings; { my $str = "0123456789"; for (1 .. 5) { chop $str; } print "[$str]"; } { my $str = "0123456789"; for (1 .. 5) { chomp $str; } print "[$str]"; } { my $str = "0123456789\n"; for (1 .. 5) { chop $str; } print "[$str]"; } { my $str = "0123456789\n"; for (1 .. 5) { chomp $str; } print "[$str]"; }` [download] It prints (observe it): `[01234][0123456789][012345][0123456789]` [download] And read perlfunc.	[reply] [d/l] [select]
Re: chop vs chomp by EvanCarroll (Chaplain) on Oct 30, 2005 at 23:12 UTC
`chop` takes off the last character of the string oblivious to what it is. `chomp` takes off $/ ("\n" by default) if it exists as the last character, also known as the input record seperater See `perldoc -f chomp`, vs `perldoc -f chop` for more information. `chop` returns what was taken off, `chomp` returns how many was taken off, (how many elements it effected). `chomp` is most often used when reading from <STDIN>. `perl -e'$_="foo\n"; chop; chomp; print;' ## Removes the , then removes the terminating $/ (\n), prints foo` `perl -e'$_="foo\n"; chomp; chop; print;' ## There is no terminating $/ (\n) chomp returns 0, chop removes , prints foo\n` Evan Carroll www.EvanCarroll.com	[reply] [d/l] [select]
Re^2: chop vs chomp by sauoq (Abbot) on Oct 30, 2005 at 23:16 UTC
Both return what was taken off. No, `chomp` returns the number of characters that were removed. As `$/` can be more than 1 character, this value is not always 0 or 1. Example: `$ perl -le '$foo = "foo\r\n"; $/ = "\r\n"; print chomp $foo; print len +gth $foo' 2 3` [download] Update: I should also mention that, when called with a list, chomp will return the sum of all characters removed from all members of the list. -sauoq "My two cents aren't worth a dime.";	[reply] [d/l] [select]