How to set octal value in java?

Question

I am trying to write the following code. but it gives me a mistake, kindly help me.

int six=06; int seven=07; int abc=018; int nine=011; System.out.println("Octal 011 ="+nine); System.out.println("octal O18 =" + abc); 

why can't i give 018 and 019 variable. I can give the value 020 and 021 variable. Why is this happening? what is the reason for this. Tell me. I got the following error

  integer number too large: 018 int eight=018; 

Answer 1

An octal number is a number system with a base of 8, that is, a digit can be from 0 to 7, you can not use the number 8 (and also 9) in the octal number system.

Answer 2

why can't I give the variable 018 and 019 a variable.

Since an integer literal with the prefix 0 treated as octal, and "8" and "9" are not valid octal digits.

From section 3.10.1 of JLS :

The octal digit consists of the ASCII digit 0, followed by one or more ASCII digits from 0 to 7, alternating with underscores, and can represent a positive, zero, or negative integer.

Trying to use “8” in an octal number is like trying to use “G” in hexadecimal ... it's just not part of the character set used in this database.

Answer 3

 // Decimal declaration and possible chars are [0-9] int decimal = 495; // HexaDecimal declaration starts with 0X or 0x and possible chars are [0-9A-Fa-f] int hexa = 0X1EF; // Octal declaration starts with 0 and possible chars are [0-7] int octal = 0757; // Binary representation starts with 0B or 0b and possible chars are [0-1] int binary = 0b111101111; 

If the number is a string format, you can convert it to int with

 String text = "0b111101111"; int value = text.toLowerCase().startsWith("0b") ? Integer.parseInt(text.substring(2), 2) : Integer.decode(text); 

Answer 4

Round numbers (base 8) can only use the following numbers: 01234567 . Just like decimal numbers (base 10) can only use 0123456789 .

So, in the octal representation, 17 + 1 is 20 .

Answer 5

why can't i give 018 and 019 variable. I can give the value 020 and 021 variable.

Leading zero means an octal literal. However, 8 and 9 are not valid octal digits. This makes errors 018 and 019 invalid.

Answer 6

The prefix 0 indicates octal (8 base) (digits 0-7).

 public class MainClass{ public static void main(String[] argv){ int intValue = 034; // 28 in decimal int six = 06; // Equal to decimal 6 int seven = 07; // Equal to decimal 7 int eight = 010; // Equal to decimal 8 int nine = 011; // Equal to decimal 9 System.out.println("Octal 010 = " + eight); } } 

Answer 7

When an integer literal starts at 0 in Java, it is considered an octal notation. Numbers 8 and 9 are illegal in octal value, numbers can vary from 0 to 7.

Answer 8

Since this is an octal number, the octal number has 8 digits that span from 0 to 7 inclusive. For the same reason, 12 would be an invalid binary number.

You need at least base 9 have 18 and a normal decimal base for 19 .

Answer 9

For your request ..... you assign an invalid value to a variable .... your assigned value starts with 0 (zero) .. which means that you assign an octal value to a variable and when you assign a value above 7, for example, 018 in your case ... the value exceeds the range of octal variables and, therefore, gives an error ... so try entering just 18 so that it is perceived as an integer, and not as a data type of an octal variable ..,