How does a leading zero change a numeric literal in Java?

Question

My friends and I have puzzled over this statement in Java after seeing it and the answer. How does this work?

System.out.printf("%d", 077);

equals 63?

oracle is here for you docs.oracle.com/javase/tutorial/java/data/numberformat.html — crAlexander
– crAlexander, Commented Feb 4, 2015 at 22:06
@David Conrad : Yes I know, I realized that after testing it in Java after receiving the answer about 077 as an Octal Number but I only asked that because it was a test questions my friends and I had and were completely confused on. — 689Shinigami
– 689Shinigami, Commented Feb 14, 2015 at 17:22
@crAlex : have already been there before I even asked the question and that does not explain anything about Octal Numbers even though I have asked about printf. — 689Shinigami
– 689Shinigami, Commented Feb 14, 2015 at 17:23

Reimeus · Accepted Answer · 2015-02-18 22:59:13Z

7

077 is an octal number which equals 7 x 8¹ + 7 x 8⁰ which is 63 decimal. To display the octal value you could do

System.out.printf("%o", 077);

answered Feb 4, 2015 at 20:56

Reimeus

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Comments

Gabriel Espinel · Accepted Answer · 2015-02-04 21:02:21Z

2

When you define an literal integer number with a 0 prefix, the compiler will treat it as an integer base 8. (Octal).

So, 77 value in Octal base is actually 63 in Decimal base.

answered Feb 4, 2015 at 21:02

Gabriel Espinel

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Vitalii Samofal · Accepted Answer · 2015-02-04 21:00:42Z

0

077 = 7 * 8^0 + 7 * 8^1 = 63; 0123 = 3 * 8^0 + 2 * 8^1 + 1 * 8^2 = 3 + 16 + 64 = 83; First 0 means octal value.

0x77 - is hex val.

answered Feb 4, 2015 at 21:00

Vitalii Samofal

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