这个正则表达式是如何工作的? [英] How does this regular expression work?

问题描述

来自 这篇文章,

/^1?$|^(11+?)1+$/ 检查一个数(它的一元值)是否是质数.

/^1?$|^(11+?)1+$/ checks whether a number(its value in unary) is prime or not.

使用这个，perl -l -e '(1 x $_) !~/^1?$|^(11+?)1+$/&&print while ++$_;' 返回素数列表.

Using this, perl -l -e '(1 x $_) !~ /^1?$|^(11+?)1+$/ && print while ++$_;' returns a list of prime numbers.

我对 Perl 没有足够的经验，但我明白正则表达式对于非质数的数字将 true.所以，如果我们用这个表达式打印所有不产生 true 的数字，我们就会得到一个质数列表.这就是 perl 查询试图做的事情.

I do not have enough experience with Perl, but what I understand is that the regular expression will be true for a number that is not prime. So, if we print all numbers that do not produce a true with this expression, we have a list of prime numbers. Thats what the perl query is trying to do.

关于正则表达式部分，

^1?$ 部分用于将 1 计数为 非质数

^1?$ part is for counting 1 as not prime

^(11+?)1+$ 用于匹配从 4 开始的非素数.

^(11+?)1+$ is for matching not prime numbers starting from 4.

我不明白为什么需要正则表达式中的 ? .据我说 /^1$|^(11+)1+$/ 应该没问题，实际上

What I do not understand is why is the ? in the regex needed at all. According to me /^1$|^(11+)1+$/ should be just fine and actually

perl -l -e '(1 x $_) !~/^1$|^(11+)1+$/&&打印 while ++$_;'给我一组相同的素数.

我对正则表达式的理解有什么缺陷吗?为什么需要 ??

? 不是应该匹配它前面的表达式的零次或一次吗?

Isn't ? supposed to match zero or one occurrence of the expression preceding it?

推荐答案

第一个 ? 用于将空字符串(即 0)匹配为非素数.如果你不在乎正则表达式是否匹配0，那就没有必要.

The first ? is for matching the empty string (i.e. 0) as non-prime. If you don't care whether the regexp matches 0, then it's not necessary.

第二个 ? 只是为了效率.+ 通常是贪婪的"，这意味着它匹配尽可能多的可用字符，然后如果正则表达式的其余部分未能匹配则回溯.+? 使它成为非贪婪的，所以它只匹配 1 个字符，然后如果正则表达式的其余部分未能匹配，则尝试匹配更多.(有关贪婪与非贪婪匹配的更多信息，请参阅 perlre 的量词部分.)

The second ? is only for efficiency. + is normally "greedy", which means it matches as many characters as are available, and then backtracks if the rest of the regexp fails to match. The +? makes it non-greedy, so it matches only 1 character, and then tries matching more if the rest of the regexp fails to match. (See the Quantifiers section of perlre for more about greedy vs. non-greedy matching.)

在这个特定的正则表达式中，(11+?) 表示它测试被 2 ('11') 和 3 ('111')，然后是 4，等等.如果你使用 (11+)，它会测试被 N(数字本身)，然后是 N-1，然后是 N-2 等的整除性.因为除数必须不大于 N/2，如果没有 ? 它将浪费时间测试许多不可能工作的潜在"除数.它仍然会匹配非质数，只是速度更慢.(此外，$1 将是最大的除数而不是最小的除数.)

In this particular regexp, the (11+?) means it tests divisibility by 2 ('11'), then 3 ('111'), then 4, etc. If you used (11+), it would test divisibility by N (the number itself), then N-1, then N-2, etc. Since a divisor must be no greater than N/2, without the ? it would waste time testing a lot of "potential" divisors that can't possibly work. It would still match non-prime numbers, just more slowly. (Also, $1 would be the largest divisor instead of the smallest one.)

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