linux 正则表达式

lzj24

求解一下：
      ^.* 这个是代表以任意多个字符开头，那么^(.*?)这个代表匹配的是什么呢？？
刚刚接触正则，不太懂...

sunzhiguolu

应该为空吧.

RE_HASH

猜不出来的时候就试试：

$> perl -e '$_='abcd'; print "Matched: -->$1<--" if (/^(.*?)/);'
Matched: --><--

jason680

回复 1# lzj24

Greedy issue


$ perldoc perlre
NAME
    perlre - Perl regular expressions

    ...

    By default, a quantified subpattern is "greedy", that is, it will match as
    many times as possible (given a particular starting location) while still
    allowing the rest of the pattern to match. If you want it to match the
    minimum number of times possible, follow the quantifier with a "?". Note
    that the meanings don't change, just the "greediness":

        *?        Match 0 or more times, not greedily
        +?        Match 1 or more times, not greedily
        ??        Match 0 or 1 time, not greedily
        {n}?      Match exactly n times, not greedily (redundant)
        {n,}?     Match at least n times, not greedily
        {n,m}?    Match at least n but not more than m times, not greedily