Why is 09 "too large" of an integer number? [duplicate]

Solution 1:

Numbers beginning with 0 are considered octal – and 9 is not an octal digit (but (conventionally) 0-7 are).

Hexadecimal literals begin with 0x, e.g. 0xA.

Up until Java 6, there was no literal notation for binary and you'll had to use something like

int a = Integer.parseInt("1011011", 2);

where the second argument specifies the desired base.

Java 7 now has binary literals.

In Java SE 7, the integral types (byte, short, int, and long) can also be expressed using the binary number system. To specify a binary literal, add the prefix 0b or 0B to the number.

Solution 2:

Integer literals beginning with a "0" are treated as octal. The permissible digits are 0 through 7.

Solution 3:

Integers beginning with the digit 0 are octal (base 8) numbers. The largest octal digit is 7; after 07 comes 010 (which is equal to decimal 8!)

012 (octal twelve) is 010 (octal ten, which is decimal 8) plus 2, or decimal 10.