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Monday, November 09, 2009

Final Thoughts: Java Puzzler: Splitting Hairs

This is the final in a series of posts about a puzzler [ post containing the question ] [ post containing the answer ]. In those two posts I highlighted some surprising behavior in String.split().

Why the surprising behavior?

Well, and here's why my job is damn cool: after discovering this issue, I dropped a note to Josh Bloch, who quickly replied: (edited summary)
Yes, this is a pain. FWIW, it was done for a very good reason: compatibility with Perl. The guy who did it is Mike "madbot" McCloskey, who now works with us at Google. Mike made sure that Java's regular expressions passed virtually every one of the 30K Perl regular expression tests (and ran faster).
I have no real issues with the way Madbot implemented regular expressions in Java, nor with the goal of Perl compatibility. Perl's regular expression language was very popular, and derivatives of it were implemented not only in Java, but , JavaScript, PCRE, Python, Ruby, Microsoft's .NET Framework, and the W3C's XML Schema.[ref]

But I do have issue with String getting saddled with a method that quietly explodes the API's complexity.

So why does Perl work this way? I don't know, dude. This isn't a Perl Puzzler.

However, I tried to recreate the original puzzler in Perl as a way to validate the original puzzler's behavior. Unfortunately, I failed using perl v5.8.8 on my OSX machine. This script:

@first = split(/:/, "");
@second = split(/:/, ":");

print scalar @first . " [@first]\n";
print scalar @second . " [@second]\n";

Yielded

0 []
0 []

I'm not claiming there's either a bug or implementation change in either the Java or Perl implementations, but I sure am curious.

Possible Solutions

1. Hacking String.split

I don't attest to this, but it seems that you can ensure consistent behavior by appending a copy of the delimiter. So, if you plan to split a string by its colons you can do:

String[] result = (string + ":").split(":")

I'm sure you can find all sorts of issues with this example. Go for it. Point them out in the comments.

Besides, that's not much of a solution.

2. Get a String tokenizer

A second solution is to use StringTokenizer, which I completely forgot about Wim Jongman made a comment in the solution post.

Of course, even it has its own specific behavior.

public class Main {
  public static void main(String[] args) {
    tokenize("");
    tokenize(":");
    tokenize("a:");
    tokenize(":a");
    tokenize("a:a");
    tokenize("::");
  }

  static void tokenize(String s) {
    StringTokenizer t = new StringTokenizer(s, ":");
    List l = new ArrayList();
    while (t.hasMoreTokens()) {
      l.add(t.nextToken());
    }
    System.out.printf("Tokenization of %s is %s\n", s, l);
  }
}

yields

Tokenization of  is []
Tokenization of : is []
Tokenization of a: is [a]
Tokenization of :a is [a]
Tokenization of a:a is [a, a]
Tokenization of :: is []

3. Get your serving of Guava.


Here's a nice one: Project Guava. Project Guava is a soon-to-be open sourced library some of Google's core Java code. I've worked with these libraries for five years and I attest that they're wonderful to use. The only problem: it's not out yet. Kevin Bourrillion tells me. though, that an initial release will be available before Thanksgiving.

Note: You may already be aware of the open source project for Google's collections library. When Guava is released, the Google Collections library will go away.

The Guava libraries have a class called Splitter. Splitter's purpose is to alleviate some the confusion that comes with String.split.
By default Splitter's behavior is very simplistic:

Splitter.on(',').split("foo,,bar,  quux")

This returns an iterable containing ["foo", "", "bar", " quux"]. Notice that the splitter does not assume that you want empty strings removed, or that you wish to trim whitespace. If you want features like these, simply ask for them:

private static final Splitter MY_SPLITTER = Splitter.on(',')
       .trimResults()
       .omitEmptyStrings();
You can read more about Guava in Kevin Bourrillion's presentation slide deck from September of this year. Splitter is covered in slides 13 to 17.

Avoiding the real issue

There are two ways of looking back on the variety of votes: either people were assuming that String.split had confusing behavior, or they just expected it to work as as they would hope. Some might want a parse of ":" to return two elements. Some might want it to return one. Or zero. Something as seemingly simple as string tokenization has behavior that just might not meet your expectations. I'd like to say that Guava's Splitter will do the trick for everyone (as it does for my case of parsing a classpath) but you need to evaluate it for yourself.

This has been a rather long way of saying: test your edge cases. Thanks for reading.

Sunday, November 08, 2009

Answer to: Java Puzzler: Splitting Hairs

Update: I fixed some small errors, and also updated the charts one last time.

This blog post contains the results and answer to the previous post, Splitting Hairs.

Given the nature of the possible answers, this is actually two puzzlers in one:
  1. How many elements come back from "".split(":"): Zero or one?
  2. How many elements come back from ":".split(":"): Zero, one or two?
Darn, I could have gotten two separate puzzlers out of this.

My Guess


Here's the important point: this is a real world case that occurred to me just the other day. Specifically, I was writing some UI code to edit and parse a colon-separated classpath, so in fact the line in question looked like this:

String[] classpathEntries = classpathField.getText().split(":");

In this case, if the UI field starts out empty, and classpathField.getText()returns an empty String. Effectively, my expectation was that with an empty classpath, I'd also get an empty array.

So my guess for "".split(":").length() is Zero.

My guess for splitting a separator would result in two empty strings on either side. So that guess is two. Put them together and you get c) 0/2.

Your Guesses

How did you folks do?

Let's look at the numbers:






An overwhelming preference for f) 1/2, and my guess, c) 0/2 in a distant second.

The Answer

If by now you haven't run the sample code, I'll tell you: the correct answer is d) 1/0. If you're shocked that you get more elements with less data, you're not alone. Hey, just look at those graphs.

Okay, let's start the analysis by reading the method signature for String.split():
public String[] split(String regex)
Ah, yes, split takes a regular expression as a delimiter, not a literal string.

Sidebar: I occasionally read about (or experience) puzzlers where someone tries splitting on a pipe character (|), which has a special meaning in regular expressions. To split on a pipe, you must use "\|" so the pattern compiler sees it as a literal character. This is not one of those puzzlers: the colon does not have a special meaning for the pattern compiler.

Part 1: "".split(":")


If you navigate through the source for String.split, you'll discover that "".split(":") is effectively the same as Pattern.compile(":").split("", 0).

Things get interesting when you read the javadoc for Pattern.split(Charsequence, int):
If this pattern does not match any subsequence of the input then the resulting array has just one element, namely the input sequence in string form.
Wha? So if the delimiter isn't found in the in the input, the original input is returned. In other words:

System.out.println(Arrays.deepToString("food".split(":")));
System.out.println(Arrays.deepToString("foo".split(":")));
System.out.println(Arrays.deepToString("fo".split(":")));
System.out.println(Arrays.deepToString("f".split(":")));
System.out.println(Arrays.deepToString("".split(":")));

yields

[food]
[foo]
[fo]
[f]
[]

OK, that explains the first one, what about the second one?

Part 2: ":".split(":")


To understand what's going on, let's look again at the javadoc for Pattern.split(). In this case, n by the nature of being called by String.split(String), is 0.
If n is zero then the pattern will be applied as many times as possible, the array can have any length, and trailing empty strings will be discarded. [ emphasis added. ]
This can be corroborated with this piece of code near the end of the method:

// Construct result
int resultSize = matchList.size();
if (limit == 0)
    while (resultSize > 0 && matchList.get(resultSize-1).equals(""))
        resultSize--;

In other words, internally, it splits the input string ":" into [ "", "" ], and then removes the elements before returning to the caller, but "".split(":") doesn't get this treatment because the delimiter was never found in the input.

Does your head hurt? Mine sure hurts, and I've done the research. But here, if you're still hungry, or hate yourself, take a look at this little gem, almost worthy of its own puzzler.

System.out.println(Arrays.deepToString((String[]) ":".split(":", 2)));
System.out.println(Arrays.deepToString((String[]) ":".split(":", 1)));
System.out.println(Arrays.deepToString((String[]) ":".split(":", 0)));

Coming Soon

Like the last puzzler, this will be followed with an analysis of why this occurs, and some nice alternatives.

Friday, November 06, 2009

Java Puzzler: Splitting Hairs

Try to answer this question without running the code, or reading the class documentation.

Assume we're using Java 1.6, though it likely doesn't matter if you're using Java 1.5.

What does the following snippet print?

String[] nothing = "".split(":");
String[] bunchOfNothing = ":".split(":");

System.out.printf("%d/%d\n", nothing.length, bunchOfNothing.length);

I will post an answer on Sunday.

This puzzler comes with six possible answers. I would have liked to have given four choices, but then I realized this was just for fun.