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I can't understand why this is not printing the expected value (400300) when I put extra zeros in front of the number:

System.out.println(new Integer(0400300)); // prints 131264
System.out.println(0400300); // prints 131264

If I put one or more zeros in front of the number, the expected value is not printed.

// JUnit test does not pass:
assertTrue(0400300 == 400300);  // returns false!?
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1 Answer 1

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10

Adding 0 to the front made the number an Octal literal. So:

0400300 = 3 * 8 ^ 2 + 4 * 8 ^ 5 = 131264

See JLS for the relevant sections. Quote:

An octal numeral consists of an ASCII digit 0 followed by one or more of the ASCII digits 0 through 7 interspersed with underscores, and can represent a positive, zero, or negative integer.

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7 Comments

It's ancient, I am not sure but I think this should have been there since day one. But with no concrete evidence (JLS 1.0) I really can't say for sure.
Hexa starts with 0x, binary with 0b (binary literals started in Java 7). But octal literals I can't find in documentation: docs.oracle.com/javase/tutorial/java/nutsandbolts/…
It's rather strange that that page mentions nothing about octal literals. Also the knowledge of binary literal is nice,
That's not documentation - it's a tutorial. You can read the link for JLS I gave in the answer and search for octal. Just edited answer too.
Well that's a bug in the tutorial. I think there are quite a handful of them, so yo could just give them a feedback by clicking the link in the bottom. But JLS is the ultimate reference in any Java related matters.
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