regex - Perl 正则表达式中的括号有什么作用？

Question

我一直在替换运算符中尝试几个正则表达式:

$str =~ s/^0+(.)/$1/;

将 0000 转换为 0 并将 0001 转换为 1

$str =~ s/^0+./$1/;

将 0000 转换为空字符串，将 000100 转换为 00，将 0001100 转换为 100。

括号有什么区别？

Answer 1

这对我来说似乎有点误用 - 您需要 () 来确定您的匹配项。

http://perldoc.perl.org/perlre.html

Capture buffers

The bracketing construct ( ... ) creates capture buffers. To refer to the current contents of a buffer later on, within the same pattern, use \1 for the first, \2 for the second, and so on. Outside the match use "$" instead of "\". (The \ notation works in certain circumstances outside the match. See the warning below about \1 vs $1 for details.) Referring back to another part of the match is called a backreference.

所以基本上你可以使用

if ($str =~ /^0+(.)/) { print "matched $1"; }

如果您有多个分组比赛，它们将是 1 美元、2 美元、3 美元...等，例如
if ($str =~ /(0*)(1*)/) { print "I've got $1 and $2"; }