匹配器在查找匹配字符时如何遍历字符串

Question

量词a?应该与a single or no occurrence of a匹配。给定的程序使用java.util.regex包将正则表达式与字符串相匹配。

我的问题是关于模式匹配的程序/结果的输出：

程序输出：-

Enter your regex: a? Enter input string to search: a I found the text "a" starting at index 0 and ending at index 1. I found the text "" starting at index 1 and ending at index 1.

问题：-

它应该匹配a的一次或零次出现。那么它不应该匹配zero-length ""(即a的不存在/零次出现(starting and ending at index 0，然后匹配a starting at index 0和ending at index 0，然后是"" starting and ending at index 1吗我认为应该。

这样一来，matcher似乎一直在字符串外寻找a，然后当它确定不再有a(即字符串的end？(时，它就会寻找zero occurrence/不存在a？我认为这会很乏味，但事实并非如此。但是，在它匹配从index 0开始到index 1结束的一个之前，它应该找到一个"starting and ending at 0？

程序：-

import java.io.InputStreamReader;
import java.util.Scanner;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
/*
 *  Enter your regex: foo
 *  Enter input string to search: foo
 *  I found the text foo starting at index 0 and ending at index 3.
 * */
public class RegexTestHarness {
    public static void main(String[] args){
        /*Console console = System.console();
        if (console == null) {
            System.err.println("No console.");
            System.exit(1);
        }*/
        while (true) {
            /*Pattern pattern = 
            Pattern.compile(console.readLine("%nEnter your regex: ", null));*/
            System.out.print("nEnter your regex: ");
            Scanner scanner = new Scanner(new InputStreamReader(System.in));
            Pattern pattern = Pattern.compile(scanner.next());
            System.out.print("nEnter your input string to seacrh: ");
            Matcher matcher = 
            pattern.matcher(scanner.next());
            System.out.println();
            boolean found = false;
            while (matcher.find()) {
                /*console.format("I found the text" +
                    " "%s" starting at " +
                    "index %d and ending at index %d.%n",
                    matcher.group(),
                    matcher.start(),
                    matcher.end());*/
                System.out.println("I found the text "" + matcher.group() + "" starting at index " + matcher.start() + " and ending at index " + matcher.end() + "."); 
                found = true;
            }
            if(!found){
                //console.format("No match found.%n", null);
                System.out.println("No match found."); 
            }
        }
    }
}

Answer 1

?量词是贪婪，这意味着它将试图找到最大可能匹配。由于匹配的部分不能重复使用，您不能在a之前匹配空字符串""(您可以将其视为第一次贪婪匹配(，但您可以在它之后匹配空字符串。

您可以通过在该量词后面添加?使其不情愿，这将使其尝试找到尽可能小的匹配。所以，如果您试图找到正则表达式a??的匹配项，您将看到0作为第一个匹配项的索引(在a之前为空字符串(。