How to include trailing empty strings in Pattern.compile("\\R").splitAsStream(input)?

user_1140754 0
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This is not a duplication of Java String split removed empty values, which deals with split() method returning a new array. In this case I would like to avoid the array.

I solved this problem with a workaround, which I am posting below as a possible solution to my question.

My goal is to process all lines including empty strings such as the following example:

String input = "foo\nbar\n\n\nzul\n\n\n";
Pattern NEWLINE = Pattern.compile("\\R");
int [] count = {1};
NEWLINE
    .splitAsStream(input)
    .forEach(line -> System.out.println(count[0]++ + ": " + line));

which produces:

1: foo
2: baz
3: 
4: 
5: zul

Yet, it is missing: 

6: 
7:

How to include last empty lines?

newline regex java
3 Answers

You can use a lookahead (?=(\\R)) to ensure that the \\R delimiter is not consumed and then remove it yourself with String.trim().

String input = "foo\nbar\n\n\nzul\n\n\n";
Pattern NEWLINE = Pattern.compile("(?=(\\R))");
int[] count = {1};
NEWLINE.splitAsStream(input)
       .map(String::trim)
       .forEach(line -> System.out.println(count[0]++ + ": " + line));

It will however result in a zero-length match for "" after the last \n.

1: foo
2: bar
3: 
4: 
5: zul
6: 
7: 
8:
user_16025550

As an alternative you can make your own implementation of an equivalent method to splitAsStream(), which includes trailing empty strings and still avoids the instantiation of an array, such as:

static Stream<String> splitAsStream(Pattern p, CharSequence input) {
    Spliterator<String> iter = new Spliterators.AbstractSpliterator<String>(
        Long.MAX_VALUE,
        Spliterator.ORDERED | Spliterator.SIZED
    ) {
        int index = 0;
        final Matcher m = p.matcher(input);

        @Override
        public boolean tryAdvance(Consumer<? super String> action) {
            while(m.find()) {
                if (index != 0 || index != m.start() || m.start() != m.end()) {
                    action.accept(input.subSequence(index, m.start()).toString());
                    index = m.end();
                    return true;
                }
            }
            if(index < input.length()) {
                // Add remaining segment
                action.accept(input.subSequence(index, input.length()).toString());
                index = input.length();
                return true;
            } else {
                return false;
            }
        }
    };
    return StreamSupport.stream(iter, false);
}
user_11407540

Since java 9 on can use Matcher.results() yielding a Stream<MatchResult>

Pattern.compile("(.*)\\R").matcher(input)
    .results(mr -> System.out.println(count[0]++ + ": " + mr.group(1)));

This guarantees that the final "line" has a terminating \n too. "....\nabc" will discard the last abc though.

For that I think the following should work (note group()), using a lookahead and requiring for the end $ at least one char ..

Pattern.compile(".*(?=\\R)|.$)").matcher(input)
    .results(mr -> System.out.println(count[0]++ + ": " + mr.group()));

A split with -1 and a check on the last entry seems a bit more readable.

user_9848230
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