How does string.split("\\S") work [duplicate]

Question

I was doing a question out of the book oracle_certified_professional_java_se_7_programmer_exams_1z0-804_and_1z0-805 by Ganesh and Sharma.

One question is:

80. Consider the following program and predict the output:

      class Test {
    
        public static void main(String args[]) {
          String test = "I am preparing for OCPJP";
          String[] tokens = test.split("\\S");
          System.out.println(tokens.length);
        }
      }
    

    a) 0

    b) 5

    c) 12

    d) 16

Now I understand that \S is a regex means treat non-space chars as the delimiters. But I was puzzled as to how the regex expression does its matching and what are the actual tokens produced by split.

I added code to print out the tokens as follows

for (String str: tokens){
  System.out.println("<" + str + ">");
}

and I got the following output

16

<>

< >

<>

< >

<>

<>

<>

<>

<>

<>

<>

<>

< >

<>

<>

< >

So a lot of empty string tokens. I just do not understand this.

I would have thought along the lines that if delimiters are non space chars that in the above text then all alphabetic chars serve as delimiters so maybe there should be 21 tokens if we are matching tokens that result in empty strings too. I just don't understand how Java's regex engine is working this out. Are there any regex gurus out there who can shed light on this code for me?

Answer 1

Copied from the API documentation: (bold are mine)

public String[] split(String regex)

Splits this string around matches of the given regular expression. This method works as if by invoking the two-argument split method with the given expression and a limit argument of zero. Trailing empty strings are therefore not included in the resulting array.

The string "boo:and:foo", for example, yields the following results with these expressions:

 Regex  Result
   :    { "boo", "and", "foo" }
   o    { "b", "", ":and:f" }

Check the second example, where last 2 "o" are just removed: the answer for your question is "OCPJP" substring is treated as a collection of separators which is not followed for non-empty strings, so that part is trimmed.

Answer 2

The reason the result is 16 and not 21 is this, from the javadoc for Split:

Trailing empty strings are therefore not included in the resulting array.

This means, for example, that if you say

"/abc//def/ghi///".split("/")

the result will have five elements. The first will be "", since it's not a trailing empty string; the others will be "abc", "", "def", and "ghi". But the remaining empty strings are removed from the array.

In the posted case:

"I am preparing for OCPJP".split("\\S")

it's the same thing. Since non-space characters are delimiters, each letter is a delimiter, but the OCPJP letters essentially don't count, because those delimiters result in trailing empty strings that are then discarded. So, since there are 15 letters in "I am preparing for", they are treated as delimiting 16 substrings (the first is "" and the last is " ").