Java ArrayList avoiding IndexOutOfBoundsException

Question

i got a short question.

ArrayList<T> x = (1,2,3,5)

int index = 6
if (x.get(6) == null) {
    return 0;
}

Exception in thread "main" java.lang.IndexOutOfBoundsException: Index: 6, Size: 4

How can I avoid this? I just want to check if there is something in the array with index 6. If there is (null)/(nothing there) i want to return 0.

You can put condition on the size of the list.

Devesh
– Devesh

2014-06-14 16:28:34 +00:00
Commented Jun 14, 2014 at 16:28 — Devesh
– Devesh, Commented Jun 14, 2014 at 16:28

Jon Skeet · Accepted Answer · 2014-06-14 16:50:43Z

18

Just use the size of the list (it's not an array):

if (x.size() <= index || x.get(index) == null) {
    ...
}

Or if you want to detect a valid index with a non-null value, you'd use:

if (index < x.size() && x.get(index) != null) {
    ...
}

In both cases, the get call will not be made if the first part of the expression detects that the index is invalid for the list.

Note that there's a logical difference between "there is no element 6" (because the list doesn't have 7 elements) and "there is an element 6, but its value is null" - it may not be important to you in this case, but you need to understand that it is different.

edited Jun 14, 2014 at 16:50

answered Jun 14, 2014 at 16:27

Jon Skeet

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4 Comments

Audrius Meškauskas Over a year ago

If x.size() == 6 then x.get(6) throws IndexOutOfBoundsException.

Jon Skeet Over a year ago

@AudriusMeškauskas: Good job that my code won't call x.get(6) in that situation then, isn't it? If x.size() == 6, then x.size() <= 6 will be true, so the second operand of the || won't be evaluated.

Audrius Meškauskas Over a year ago

Still looks very twisted. Your own second example is much easier to read.

Jon Skeet Over a year ago

@AudriusMeškauskas: The second example is the inverse though - look at the OP's original code; he's trying to detect the absence of a non-null value. That's the case if either the index is out of range or the result is null. Yes, it could be written as if (index >= x.size() ...) if you'd prefer that, but it would still fundamentally be a condition of "if the index is out of range" vs the second snippet of "if the index is in range".

Ashish Bhatia · Accepted Answer · 2014-06-14 16:29:16Z

2

check size of arraylist. if size is less that 6 then return 0 else return value

answered Jun 14, 2014 at 16:29

Ashish Bhatia

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1 Comment

Lluis Martinez Over a year ago

Less or equal (off by one).

Rakesh KR · Accepted Answer · 2014-06-14 16:35:42Z

0

Check the size of the list first by using size(), then check for the index.

if (x.size() <= 6 || x.get(6) == null) {
    return 0;
}

answered Jun 14, 2014 at 16:35

Rakesh KR

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