[ACCEPTED]-In Perl, what is the difference between s/^\s+// and s/\s+$//?-perl
From perldoc perlfaq4:
How do I strip blank space from the beginning/end of a string?
A substitution can do this for you. For 22 a single line, you want to replace all 21 the leading or trailing whitespace with 20 nothing. You can do that with a pair of 19 substitutions:
s/^\s+//; s/\s+$//;You can also write that as 18 a single substitution, although it turns out 17 the combined statement is slower than the 16 separate ones. That might not matter to 15 you, though:
s/^\s+|\s+$//g;In this regular expression, the 14 alternation matches either at the beginning 13 or the end of the string since the anchors 12 have a lower precedence than the alternation. With 11 the
/gflag, the substitution makes all 10 possible matches, so it gets both. Remember, the 9 trailing newline matches the\s+, and the 8$anchor can match to the absolute end 7 of the string, so the newline disappears 6 too.
And from perldoc perlrequick:
To specify where it should 5 match, we would use the anchor metacharacters 4
^and$. The anchor^means match at the beginning 3 of the string and the anchor$means match 2 at the end of the string, or before a 1 newline at the end of the string. Some examples:"housekeeper" =~ /keeper/; # matches "housekeeper" =~ /^keeper/; # doesn't match "housekeeper" =~ /keeper$/; # matches "housekeeper\n" =~ /keeper$/; # matches "housekeeper" =~ /^housekeeper$/; # matches
^ means starts with, $ means ends with this 1 string.
The first one will only replace whitespace 1 at the beginning of the line.
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