Strange array behavior in Java

Question

This is a simple array declaration and initialization.

  int arr[] = new int[10];

    for(int i = 0; i<arr.length; i++){

    arr[i] = i;
    }

This

System.out.println(arr[000001]);

to

System.out.println(arr[000007]);

prints out the correct values but anything above 8

System.out.println(arr[000008]);

produces a java.lang.RuntimeException: Uncompilable source code

Why does this happen?

Never saw that exception in my life. In my IDE it just doesn't compile. — Mister Smith
– Mister Smith, Commented Sep 30, 2011 at 12:12
Allright, it is NetBeans stuff. I was surprised of that exception being thrown when running an already compiled application! — Mister Smith
– Mister Smith, Commented Sep 30, 2011 at 12:15

Ernest Friedman-Hill · Accepted Answer · 2011-09-30 11:35:07Z

13

This has nothing to do with arrays; integers starting with the digit 0 are octal (base 8). The legal octal digits are 0-7, so that 08 (or 00000008) are invalid octal integer literals. The correct octal for 8 is 010.

answered Sep 30, 2011 at 11:35

Ernest Friedman-Hill

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S.L. Barth is on codidact.com · Accepted Answer · 2011-09-30 11:36:00Z

6

It's because the 0's in front of your index make Java think you're using the octal numbering system.

answered Sep 30, 2011 at 11:36

S.L. Barth is on codidact.com

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Joachim Sauer · Accepted Answer · 2011-09-30 11:35:47Z

5

It has nothing to do with arrays.

Integer literals that start with a 0 are expected to be octal numerals.

Therefore, if you have any diggit bigger than 7 (i.e. 8 or 9) in there, then it won't compile.

Also: you only get an Exception because your IDE allows you to execute code that doesn't compile. That's a very bad idea, you should look at the compiler error it produces instead (it will probably have much more information than the message you posted).

answered Sep 30, 2011 at 11:35

Joachim Sauer

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nos · Accepted Answer · 2011-09-30 11:36:57Z

4

It happens because 000001 , 000007, 000008 is octal notation. Integer literals starting with 0 is treated as octal. However there is no such thing as 000008 in a base 8 numeral system (octal).

(Though, I would have expected that to fail during compile time, not runtime)

answered Sep 30, 2011 at 11:36

nos

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