这个替换有什么问题吗？
#!/usr/bin/perl
use strict;
use warnings;
my $s = "sad day
Good day
May be Bad Day
";
$s =~ s/\w+ \w+/_/gm;
print $s;
我试图用 _ 替换单词之间的所有空格，但它不起作用。这有什么问题吗？
#!/usr/bin/perl
use strict;
use warnings;
my $s = "sad day
Good day
May be Bad Day
";
$s =~ s/\w+ \w+/_/gm;
print $s;
I am trying to substitute all spaces between words with _, but it is not working. What is wrong with that?
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替换会替换整个单词 (
\w+)，然后替换一个空格，然后用下划线替换另一个单词。实际上没有必要替换（或捕获重要的内容）这些单词
将替换分词符（
\b，单词和非单词之间的零宽度转换），后跟一个或多个空格后跟另一个分词符和下划线。使用\b可确保您不会替换新行前后的空格。The substitution replaces an entire word (
\w+) then a space then an other word by an underscore.There is really no need to replace (or capture for what matters) those words
will replace a word break (
\b, a zero-width transition between a word and a non word) followed by one or more spaces followed by an other word break, by an underscore. Using\bensures that you don't replace a space after or before a new line.这种模式替换可能是最有效的解决方案：
This pattern replacement is probably the most efficient solution:
如果没有涉及明确 look- 的答案，这个问题就不完整前向和后向断言：
这实际上与涉及零宽度字边界锚点
\b的解决方案相同。请注意，前向正则表达式可以匹配任何字符长度的正则表达式，但后向正则表达式必须是固定长度的（这就是为什么
(?<=\w+)可以'This question wouldn't be complete without an answer involving explicit look-ahead and look-behind assertions:
This is effectively the same as the solutions involving the zero-width word-boundary anchor,
\b.Note that look-ahead regexes can match regexes of any character length, but look-behind regexes have to be of fixed-length (which is why
(?<=\w+)can't be done).