理解所有格量词，java regex [英] Understanding possessive quantifiers, java regex

问题描述

我知道占有正则表达式会转到文本的末尾，并且不会回溯以查看结尾之前是否有匹配项.如果最后有匹配项，则返回 true，否则立即返回 false.我已经试过了:

I understand that a possesive regex would go to the end of the text and would not backtrack to see if there was a match before the end. If at the end there's a match it returns true, otherwise it immidiatly returns false. I've tride this:

Pattern patt = Pattern.compile(".*+foo");
Matcher matcher = patt.matcher("xxfooxxxxxfooxxxfoo");
while (matcher.find())
    System.out.println(matcher.group());

即使最后有一场比赛，它也没有给我任何东西.任何想法为什么?

It gives me nothing even though there's a match in the end. Any ideas why?

我也明白要使正则表达式变得懒惰/占有，我在第一个量词(即 *? 或 *+)之后添加 ?/+ .是对的吗?谢谢！

Also I understand that to make a regex lazy/possessive I add ?/+ after the first quantifier (i.e. *? or *+). Is that right? Thanks!

推荐答案

即使最后有一场比赛，它也没有给我任何东西.任何想法为什么?

It gives me nothing even though there's a match in the end. Any ideas why?

.*+ 将匹配整个输入字符串(包括最后一个 foo).并且因为它不会从字符串的末尾回溯，所以正则表达式 .*+foo 不匹配.

The .*+ will match the entire input string (including the last foo). And because it does not backtrack from the end of the string, the regex .*+foo does not match.

我也明白要使正则表达式变得懒惰/占有，我在第一个量词(即 *? 或 *+)之后添加 ?/+ .是吗?

Also I understand that to make a regex lazy/possessive I add ?/+ after the first quantifier (i.e. *? or *+). Is that right?

所有格的反义词是不懒惰.这将是贪婪的，默认情况下 * 是.

The counter part of possessive is not lazy. That would be greedy, which * by default is.

因此，正则表达式 .*?foo 将匹配 "xxfoo" 而正则表达式 .*foo 将匹配 "xxfooxxxxxfooxxxfoo".

So, the regex .*?foo would match "xxfoo" and the regex .*foo would match "xxfooxxxxxfooxxxfoo".

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