What does an integer that has zero in front of it mean and how can I print it?

Question

class test{
  public static void main(String args[]){
     int a = 011;
     System.out.println(a);
  }
}

Why I am getting 9 as output instead of 011?

How can I get 011 as output?

Unsnap

String.format("%02d", a) – Gipps

011 is a literal in base 8, so in base 10 you get 9=(8^1+8^0). see https://mcmap.net/q/47819/-why-is-08-not-a-valid-integer-literal-in-java for more details – Marquisette

numbers that start with 0 are called hex numbers. – Priory

@Priory not exactly. Numbers written with a leading "0" in Java are called octal and are base 8 (octa- is the Greek prefix for 8). Numbers written with a leading "0x" are called hexadecimal (hex for short) and are base 16 (hexa- is the Greek prefix for 6 and deci- is the Latin prefix for 10). Regular numbers are called decimal because they are base 10. Octal and hexadecimal are useful because they are much easier to convert to binary than decimal. – Bespeak

Answer 1

The JLS 3.10.1 describes 4 ways to define integers.

An integer literal may be expressed in decimal (base 10), hexadecimal (base 16), octal (base 8), or binary (base 2).

An octal numeral consists of a digit 0 followed by one or more of the digits 0 through 7 ...

A decimal numeral is either the single digit 0, representing the integer zero, or consists of an digit from 1 to 9 optionally followed by one or more digits from 0 to 9 ...

In summary if your integer literal (i.e. 011) starts with a 0, then java will assume it's an octal notation.

Solutions:

If you want your integer to hold the value 11, then don't be fancy, just assign 11. After all, the notation doesn't change anything to the value. I mean, from a mathematical point of view 11 = 011 = 11,0.

int a = 11;

The formatting only matters when you print it (or when you convert your int to a String).

String with3digits = String.format("%03d", a);
System.out.println(with3digits);

The formatter "%03d" is used to add leading zeroes.

Alternatively, you could do it in 1 line, using the printf method.

System.out.printf("%03d", a);

Turves

Answer 2

A numeric literal that starts with 0 is parsed as an octal number (i.e. radix 8). 011 is 9 in octal.

Yser

Answer 3

The 011 is being interpreted as an octal number. This means it's in base 8. See also this SO post. From the accepted answer by Stuart Cook:

In Java and several other languages, an integer literal beginning with 0 is interpreted as an octal (base 8) quantity.

Digits in Base 10:
100s, 10s, 1s

Digits in Base 8:
64s, 8s, 1s

So the 011 is interpreted as 8+1=9

Stines