在Perl中，s / ^ \\ s + //和s / \\ s + $ //之间有什么区别？

Question

I know that the following three lines of codes aim to extract the string into $value and store it in $header. But I do not know what are the differences between $value =~ s/^\\s+//; and $value =~ s/\\s+$//; .

$value =~ s/^\s+//;
$value =~ s/\s+$//;
$header[$i]= $value;

Answer 1

From perldoc perlfaq4 :

How do I strip blank space from the beginning/end of a string?

A substitution can do this for you. For a single line, you want to replace all the leading or trailing whitespace with nothing. You can do that with a pair of substitutions:

 s/^\\s+//; s/\\s+$//; 

You can also write that as a single substitution, although it turns out the combined statement is slower than the separate ones. That might not matter to you, though:

 s/^\\s+|\\s+$//g; 

In this regular expression, the alternation matches either at the beginning or the end of the string since the anchors have a lower precedence than the alternation. With the /g flag, the substitution makes all possible matches, so it gets both. Remember, the trailing newline matches the \\s+ , and the $ anchor can match to the absolute end of the string, so the newline disappears too.

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And from perldoc perlrequick :

To specify where it should match, we would use the anchor metacharacters ^ and $ . The anchor ^ means match at the beginning of the string and the anchor $ means match at the end of the string, or before a 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 

Answer 2

^表示开头，$表示以此字符串结尾。

Answer 3

第一个只会替换行开头的空格。