Java 在单行上解析块注释

我想忽略/瞄准线的非块注释段

例如，以下字符串需要全部生成一个字符串

“foobar”

---

实现此功能的最佳方法是什么？

编辑请尝试以下方法：

public static void main(String[] args) {

    String[] input = new String[]{"foo/*comment*/bar", "comm*/foobar/*ent", "comment*/foobar", "foobar/*comment"};
    String pattern = "(?:/\\*[^\\*]+(?:\\*/)?|(?:/\\*)?[^\\*]+\\*/)";

    List<String> listMatches = new ArrayList<String>();
    String result = "";
    for (String m : input) {
        result = m.replaceAll(pattern, ""); //remove matches
        listMatches.add(result); // append to list
        System.out.println(result);
    }
}

以下是正则表达式的解释：

(?:         1st non-capturing group starts
/\\*        match /* literally
[^\\*]+     1 or more times characters except *
(?:         2nd non-capturing group starts
\\*/        match */ literally      
)           2nd non-capturing group ends
?           match previous non-capturing group 0 or 1 time
|           Or (signals next alternative)
(?:         3rd non-capturing group starts
/\\*        match /* literally
)           3rd non-capturing group ends
?           match previous non-capturing group 0 or 1 time
[^\\*]+     1 or more times characters except *
\\*/        match */ one time
)           1st non-capturing group ends    

---

这与post in具有相同的逻辑，但以递归形式实现，以满足您对简单性的需求：

public static String cleanComment(String str) {
    int open = str.indexOf("/*"), close = str.indexOf("*/");

    if( (open&close) < 0 ) return str;

    open &= Integer.MAX_VALUE;
    close &= Integer.MAX_VALUE;

    if(open < close) {
        if(close > str.length()) {
            return str.substring(0, open);
        } else { 
            return str.substring(0, open) + cleanComment( str.substring(close+2) );
        }
    } else {
        return cleanComment( str.substring(close+2) );
    }       
}

publicstaticstringcleancoment（stringstr）{
int open=str.indexOf（“/”），close=str.indexOf（“*/”）；
如果（（打开和关闭）<0）返回str；
打开&=Integer.MAX_值；
关闭&=Integer.MAX_值；
如果（打开<关闭）{
如果（关闭>str.length（））{
返回str.substring（0，打开）；
}否则{
返回str.substring（0，打开）+cleanComment（str.substring（关闭+2））；
}
}否则{
返回cleanComment（str.substring（close+2））；
}       
}

看看这个解决方案似乎相当复杂。我正试图找出最简单的解决办法。这个问题怎么会太宽泛了。这是一个目标明确的简单问题。我希望这些内容不在注释中。@Ogen:由于正则表达式正在查找所有注释，我们只需要删除所有匹配项。这实际上很有效。你能解释一下正则表达式的不同部分在做什么吗？上面的解释对你有好处吗？是的，很好，谢谢

(?:         1st non-capturing group starts
/\\*        match /* literally
[^\\*]+     1 or more times characters except *
(?:         2nd non-capturing group starts
\\*/        match */ literally      
)           2nd non-capturing group ends
?           match previous non-capturing group 0 or 1 time
|           Or (signals next alternative)
(?:         3rd non-capturing group starts
/\\*        match /* literally
)           3rd non-capturing group ends
?           match previous non-capturing group 0 or 1 time
[^\\*]+     1 or more times characters except *
\\*/        match */ one time
)           1st non-capturing group ends    

public static String cleanComment(String str) {
    int open = str.indexOf("/*"), close = str.indexOf("*/");

    if( (open&close) < 0 ) return str;

    open &= Integer.MAX_VALUE;
    close &= Integer.MAX_VALUE;

    if(open < close) {
        if(close > str.length()) {
            return str.substring(0, open);
        } else { 
            return str.substring(0, open) + cleanComment( str.substring(close+2) );
        }
    } else {
        return cleanComment( str.substring(close+2) );
    }       
}