Removing all punctuation using POSIX in Java and C# produce different output

Question

Here is my try:

Java:

public static void main(String[] args) {
 String text = "This && is **^^ a ~~@@ test.";
 System.out.println(Pattern.compile("\\p{Punct}").matcher(text).replaceAll(""));
 // OUT: This  is  a  test --> As I expected
}

C#:

static void Main(string[] args) {
 string text = "This && is **^^ a ~~@@ test.";
 Console.WriteLine(Regex.Replace(text, "\\p{P}", ""));
 // OUT: This  is ^^ a ~~ test
 // expected: This  is  a  test
 Console.ReadLine();
}

Any ideas? Thank you!

See also: [Are Java and C# regular expressions compatible?](https://stackoverflow.com/questions/538579/are-java-and-c-sharp-regular-expressions-compatible). (Spoiler alert: not necessarily...) — paulsm4, Dec 03 '18 at 19:56

score 2 · Accepted Answer · answered Dec 03 '18 at 17:57

2

"\\p{P}" means that same in both Java and C#, i.e. match Unicode Category P (Punctuation).

Java's "\\p{Punct}" means something else, and is documented as:

Punctuation: One of !"#$%&'()*+,-./:;<=>?@[\]^_`{|}~

So, the equivalent C# is "[!\"#$%&'()*+,\\-./:;<=>?@\\[\\\\\\]^_`{|}~]"

answered Dec 03 '18 at 17:57

Andreas

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Vote! I didn't know \\p{P} works in Java. My question is what is equivalent of \\p{Punct} in C# in POSIX style? – LHA Dec 03 '18 at 18:12
1

@Loc C# regex does not support POSIX character classes, so use the character class Andreas shared. – Wiktor Stribiżew Dec 03 '18 at 18:15
1

@Loc C# isn't POSIX compliant so it doesn't provide POSIX character classes. – revo Dec 03 '18 at 18:15
Thank you for letting me know that! – LHA Dec 03 '18 at 18:21
1

@Loc I already told you what the equivalent of ``\\p{Punct}`` is in C#. Please read last line of answer. – Andreas Dec 03 '18 at 18:26

Removing all punctuation using POSIX in Java and C# produce different output

1 Answers1