正则表达式在Java中没有明显的最大长度 [英] Regex look-behind without obvious maximum length in Java
问题描述
我一直认为Java的regex-API（以及许多其他语言）中的 look-behind 断言必须具有明显的长度。因此，不允许在 look-behinds 中使用STAR和PLUS量词。
I always thought that a look-behind assertion in Java's regex-API (and many other languages for that matter) must have an obvious length. So, STAR and PLUS quantifiers are not allowed inside look-behinds.
优秀的在线资源 regular-expressions.info 似乎证实（某些）我的假设：
The excellent online resource regular-expressions.info seems to confirm (some of) my assumptions:
[...] Java通过
允许有限重复更进一步。你仍然
不能使用星号或加号，但你
可以使用问号和
花括号，指定了max参数
.Java识别
有限重复可以
重写为字符串
的替换，具有不同但固定的长度。
不幸的是，当你在lookbehind中使用
替换时，JDK 1.4和1.5
有一些错误。这些
在JDK 1.6中修复。[...]
"[...] Java takes things a step further by allowing finite repetition. You still cannot use the star or plus, but you can use the question mark and the curly braces with the max parameter specified. Java recognizes the fact that finite repetition can be rewritten as an alternation of strings with different, but fixed lengths. Unfortunately, the JDK 1.4 and 1.5 have some bugs when you use alternation inside lookbehind. These were fixed in JDK 1.6. [...]"
- http://www.regular-expressions.info /lookaround.html
只要前瞻中的字符范围的总长度小于或等于Integer.MAX_VALUE，就使用大括号。所以这些正则表达式是有效的：
Using the curly brackets works as long as the total length of range of the characters inside the look-behind is smaller or equal to Integer.MAX_VALUE. So these regexes are valid:
"(?<=a{0," +(Integer.MAX_VALUE) + "})B"
"(?<=Ca{0," +(Integer.MAX_VALUE-1) + "})B"
"(?<=CCa{0," +(Integer.MAX_VALUE-2) + "})B"
但这些不是：
"(?<=Ca{0," +(Integer.MAX_VALUE) +"})B"
"(?<=CCa{0," +(Integer.MAX_VALUE-1) +"})B"
但是，我不明白以下内容：
However, I don't understand the following:
当我使用后瞻中的*和+量词运行测试时，一切顺利（见输出测试1 和测试2 ）。
When I run a test using the * and + quantifier inside a look-behind, all goes well (see output Test 1 and Test 2).
但是，当我添加一个字符时在测试1 和测试2 的后视开始时，它会中断（请参阅输出测试3 ）。
But, when I add a single character at the start of the look-behind from Test 1 and Test 2, it breaks (see output Test 3).
从测试3 不情愿的贪婪*没有效果，它仍然会中断（参见测试4 ）。
Making the greedy * from Test 3 reluctant has no effect, it still breaks (see Test 4).
这里是测试工具：
public class Main {
private static String testFind(String regex, String input) {
try {
boolean returned = java.util.regex.Pattern.compile(regex).matcher(input).find();
return "testFind : Valid -> regex = "+regex+", input = "+input+", returned = "+returned;
} catch(Exception e) {
return "testFind : Invalid -> "+regex+", "+e.getMessage();
}
}
private static String testReplaceAll(String regex, String input) {
try {
String returned = input.replaceAll(regex, "FOO");
return "testReplaceAll : Valid -> regex = "+regex+", input = "+input+", returned = "+returned;
} catch(Exception e) {
return "testReplaceAll : Invalid -> "+regex+", "+e.getMessage();
}
}
private static String testSplit(String regex, String input) {
try {
String[] returned = input.split(regex);
return "testSplit : Valid -> regex = "+regex+", input = "+input+", returned = "+java.util.Arrays.toString(returned);
} catch(Exception e) {
return "testSplit : Invalid -> "+regex+", "+e.getMessage();
}
}
public static void main(String[] args) {
String[] regexes = {"(?<=a*)B", "(?<=a+)B", "(?<=Ca*)B", "(?<=Ca*?)B"};
String input = "CaaaaaaaaaaaaaaaBaaaa";
int test = 0;
for(String regex : regexes) {
test++;
System.out.println("********************** Test "+test+" **********************");
System.out.println(" "+testFind(regex, input));
System.out.println(" "+testReplaceAll(regex, input));
System.out.println(" "+testSplit(regex, input));
System.out.println();
}
}
}
输出：
********************** Test 1 **********************
testFind : Valid -> regex = (?<=a*)B, input = CaaaaaaaaaaaaaaaBaaaa, returned = true
testReplaceAll : Valid -> regex = (?<=a*)B, input = CaaaaaaaaaaaaaaaBaaaa, returned = CaaaaaaaaaaaaaaaFOOaaaa
testSplit : Valid -> regex = (?<=a*)B, input = CaaaaaaaaaaaaaaaBaaaa, returned = [Caaaaaaaaaaaaaaa, aaaa]
********************** Test 2 **********************
testFind : Valid -> regex = (?<=a+)B, input = CaaaaaaaaaaaaaaaBaaaa, returned = true
testReplaceAll : Valid -> regex = (?<=a+)B, input = CaaaaaaaaaaaaaaaBaaaa, returned = CaaaaaaaaaaaaaaaFOOaaaa
testSplit : Valid -> regex = (?<=a+)B, input = CaaaaaaaaaaaaaaaBaaaa, returned = [Caaaaaaaaaaaaaaa, aaaa]
********************** Test 3 **********************
testFind : Invalid -> (?<=Ca*)B, Look-behind group does not have an obvious maximum length near index 6
(?<=Ca*)B
^
testReplaceAll : Invalid -> (?<=Ca*)B, Look-behind group does not have an obvious maximum length near index 6
(?<=Ca*)B
^
testSplit : Invalid -> (?<=Ca*)B, Look-behind group does not have an obvious maximum length near index 6
(?<=Ca*)B
^
********************** Test 4 **********************
testFind : Invalid -> (?<=Ca*?)B, Look-behind group does not have an obvious maximum length near index 7
(?<=Ca*?)B
^
testReplaceAll : Invalid -> (?<=Ca*?)B, Look-behind group does not have an obvious maximum length near index 7
(?<=Ca*?)B
^
testSplit : Invalid -> (?<=Ca*?)B, Look-behind group does not have an obvious maximum length near index 7
(?<=Ca*?)B
^
我的问题可能很明显，但我仍会问：任何人都可以向我解释为什么测试1 和 2 失败，而测试3 和 4 不会？我本以为失败，其中一半没有工作，一半失败。
My question may be obvious, but I'll still ask it: Can anyone explain to me why Test 1 and 2 fail, and Test 3 and 4 don't? I would have expected them all to fail, not half of them to work and half of them to fail.
谢谢。
PS。我正在使用：Java版本1.6.0_14
PS. I'm using: Java version 1.6.0_14
推荐答案
浏览Pattern.java的源代码会发现'*'并且'+'被实现为Curly的实例（这是为卷曲运算符创建的对象）。所以，
Glancing at the source code for Pattern.java reveals that the '*' and '+' are implemented as instances of Curly (which is the object created for curly operators). So,
a*
实现为
a{0,0x7FFFFFFF}
和
a+
实现为
a{1,0x7FFFFFFF}
这就是为什么你会看到与curlies和stars完全相同的行为。
which is why you see exactly the same behaviors for curlies and stars.
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