英文:
A problem about the code in java especially in skip function
问题
scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
can anyone please explain this code...
i'm still confusing about this code and how this code works while compiling
英文:
scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
can anyone please explain this code...
i'm still confusing about this code and how this code works while compiling
答案1
得分: 1
scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
你可以阅读预定义字符类的所有详细信息：
https://docs.oracle.com/javase/10/docs/api/java/util/regex/Pattern.html
查找水平空白和换行匹配器。
英文:
scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
You can read all the details of predefined character classes:
https://docs.oracle.com/javase/10/docs/api/java/util/regex/Pattern.html
Look for horizontal white space and linebreak matcher.
答案2
得分: 0
这段代码告诉扫描器在输入中跳过不需要的字符，忽略分隔符。
在这种情况下，一组与换行处理相关的字符会被跳过：
- 一对
\r\nCRLF，如Windows中所示 - 或以下任何单个字符：
\r- CR（回车）MacOS换行分隔符\n- LF（换行）Unix换行分隔符\u2028- Unicode换行分隔符\u2029- Unicode段落分隔符\u0085- Unicode下一行
英文:
This code tells scanner to skip unwanted characters in the input ignoring delimiters.
In this case, a set of characters related to new-line handling are skipped:
- a pair of
\r\nCRLF as in Windows - or any of the following single characters:
\r- CR (carriage-return) MacOS line separator\n- LF (line-feed) Unix line separator\u2028- Unicode line separator\u2029- Unicode paragraph separator\u0085- Unicode next line
答案3
得分: 0
Explanation for the string:
\r\n - 一对CRLF，类似于Windows中的换行符，
\r - CR（回车）MacOS换行符，
\n - LF（换行）Unix换行符，
\u2028 - Unicode换行符，
\u2029 - Unicode段落分隔符，
\u0085 - Unicode下一行。
英文:
Explanation for the string:
\r\n - a pair of CRLFs like in Windows, or any of the following single characters:
\r - CR (carriage return) MacOS line separator,
\n - LF (line feed) Unix line separator,
\u2028 - Unicode line separator,
\u2029 - Unicode paragraph separator,
\u0085 - Unicode next line.
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