CODEWITHSUNDEEP
java
pankaj4u4m
pankaj4u4m

Reputation: 327

why split function behave like this in java?

if I do 

    String a = ""
    String b = a.split(" ")[0];

It is not giving ArrayIndexOutOfBoundException

but when I do 

    String a = " "
    String b = a.split(" ")[0];

It is giving me ArrayIndexOutOfBoundException

again when I do

    String a = " abc"
    String b = a.split(" ")[0];

It is not giving me Exception WHY SO?

Upvotes: 11

Views: 231

Answers (5)

Kush
Kush

Reputation: 148

String a = "";

String b = a.split(" ")[0];

When you do this, since there is no 'split' character present, therefore no split action is performed and an array is returned with first and only element as an empty string.

String a = " ";

String b = a.split(" ")[0];

while in this case it tries to split the string but get no values to be placed in either side, therefore no array is created. thus when you are trying to access its 0th element, it is giving ArrayOutOfBoundException.

String a = " abc";

String b = a.split(" ")[0];

In this case, splitting takes place and "abc" is placed at 0th place (I guess) and you are left with an array with a size greater that 0. Problem Solved!!..

Upvotes: 1

NPE
NPE

Reputation: 500357

Case #1

String a = ""
String b = a.split(" ")[0];

From the Javadoc:

If the expression does not match any part of the input then the resulting array has just one element, namely this string.

So split() gives you a single-element array, consisting of the input string (""). Therefore there's no exception.

Case #2

String a = " "
String b = a.split(" ")[0];

From the Javadoc:

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

You have two such trailing empty strings, and nothing else. Therefore you get back a zero-size array, resulting in the exception.

Case #3

String a = " abc"
String b = a.split(" ")[0];

It's not much of a corner case as it clearly has to return at least one element, hence no exception.

Upvotes: 2

Harry Lime
Harry Lime

Reputation: 29576

It's always a possibility that String.split() will return an empty array if there is a complete match on the provided regex

Upvotes: 0

pcalcao
pcalcao

Reputation: 15990

It's kind of weird.

Thing is, in your first example, the empty String "" is a String, not null. So when you say: Split this "" with the token " ", the patterns doesn't match, and the array you get is the original String. Same as if you did 

String a = "abc";
String b = a.split(" ")[0];

The pattern to split doesn't match, so you get one token, the original String.

You get an exception on the second case because the COMPLETE content of your String is exactly the delimiter you've passed to split, so you end up with an empty array.

Let me know if you want some further details, but this is pretty much it.

Upvotes: 3

PaoloVictor
PaoloVictor

Reputation: 1306

After:

String a = " "
String[] arr = a.split(" ");

arr is an empty array. That's why it throws an ArrayIndexOutOfBoundsException when you try to access its first (and non-existant) element. Now, after:

String a = " abc"
String[] arr = a.split(" ");

arr has one element: "abc", which is why no exception is thrown when you try to access its first element.

Upvotes: 2

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