ExamLab: invalid expression String [] s = files.split("[(\\s)+]");

[Harry Henriques]

Hi,

This expression was listed as one of the correct choices in ExamLab for a String.split() delimiter where one or more whitespace characters separate the tokens. The brackets [ ] denote a regex character class.

I think the correct expressions should be one or both of the following:

String [ ] s = files.split("[\\s]+");
String [ ] s = files.split("\([\\s]+\)");    

What do you think?

Harry

The Java Tutorials on Regular Expressions wrote:A regex character class is a set of characters enclosed within square brackets. It specifies the characters that will successfully match a single character from a given string

Pattern.html wrote:
Character classes
[abc] a, b, or c (simple class)
[^abc] Any character except a, b, or c (negation)
[a-zA-Z] a through z or A through Z, inclusive (range)
[a-d[m-p]] a through d, or m through p: [a-dm-p] (union)
[a-z&&[def]] d, e, or f (intersection)
[a-z&&[^bc]] a through z, except for b and c: [ad-z] (subtraction)
[a-z&&[^m-p]] a through z, and not m through p: [a-lq-z](subtraction)

[Harsh Pensi]

Harry Henriques wrote:Hi,

I think the correct expressions should be one or both of the following:

String [ ] s = files.split("[\\s]+");
String [ ] s = files.split("\([\\s]+\)");    

What do you think?

The given regex [(\\s)+] does not work as desired.
1. is correct. 2. has invalid escape sequence

[Harry Henriques]

You are correct, Harsh. The following has an invalid escape sequence.

String [ ] s = files.split("\([\\s]+\)");

How about this?

String files = "The quick brown    fox";
String [ ] s = files.split("\\([\\s]\\)+");
for( String b : s )
    System.out.print( b + " " );
System.out.println ();

or this?

String files = "The quick brown    fox";
String [ ] s = files.split("\\([\\s]+\\)");
for( String b : s )
    System.out.print( b + " " );
System.out.println ();

The output for the previous 2 snippets is:

The quick brown   fox // with 3 spaces between "brown" and "fox"

String files = "The quick brown     fox";
String [ ] s = files.split("[\\s]+");
for( String b : s )
    System.out.print( b + " " );
System.out.println ();

The output for the last one is:

The quick brown fox // with 1 space between "brown" and "fox"

Do you see the difference? Why does it do this?

Regards, Harry

[Henry Wong]

Do you see the difference? Why does it do this?

In the first two cases, nothing is split, as there are no delimiters that has an open paren, a bunch of spaces, and a close paren -- split returns an array of size one, which contains the original string.

In the last case, the delimiter is just one or more spaces, so it does split it into a bunch of words.


Oops... Small correction. Only the second example's delimiter is an open parens, a bunch of spaces, and a close paren. The first example's delimiter is an open paren, one space, and a bunch of close parens. Regardless, neither case will match a delimiter for the string, and nothing will be split.

Henry

[Harry Henriques]

Thanks, Henry. I see my mistake, now. It is not necessary to escape the parenthesis.

String files = "The quick brown    fox";   
String [ ] s = files.split("([\\s])+");   
for( String b : s )   
    System.out.print( b + " " );   
System.out.println ();  

String files = "The quick brown    fox";   
String [ ] s = files.split("([\\s]+)");   
for( String b : s )   
    System.out.print( b + " " );   
System.out.println ();  

Both of the above regular expressions split the String into one word tokens. The result of both of these is the following.

The quick brown fox // with 1 space between "brown" and "fox"

Regards, Harry