CODEWITHSUNDEEP
javanumbersintegerliterals
Akash Thakare
Akash Thakare

Reputation: 23012

Why int j = 012 giving output 10?

In my actual project It happened accidentally here is my modified small program.

I can't figure out why it is giving output 10?

public class Int
{
    public static void main(String args[])
    {
        int j=012;//accidentaly i put zero 
        System.out.println(j);// prints 10??
    }
}

After that, I put two zeros still giving output 10.

Then I change 012 to 0123 and now it is giving output 83?

Can anyone explain why?

Upvotes: 14

Views: 27966

Answers (5)

Monu Rohilla
Monu Rohilla

Reputation: 665

By placing a zero in front of the number is an integer in octal form. 010 is in octal form so its value is 8

Guess output of this program on your own :

 public class sol{
       public static void main(String[] args){
             int p = 010;
             int q = 06;
             System.out.println(p);
             System.out.println(q);
       }
 }

output is :

8
6

Upvotes: 1

Amit
Amit

Reputation: 72

If a number is leading with 0, the number is interpreted as an octal number, not decimal. Octal values are base 8, decimal base 10.

System.out.println(012):
(2 * 8 ^ 0) + (1 * 8 ^ 1) = 10
12 = 2*8^0 + 1*8^1 ---> 10


System.out.println(0123)
(3 * 8 ^ 0) + (2 * 8 ^ 1) + (1 * 8 ^ 2) = 83
123 = 3*8^0 + 2*8^1 + 1*8^2 ---> 83

Upvotes: 3

Abimaran Kugathasan
Abimaran Kugathasan

Reputation: 32498

Than I change 012 to 0123 and now it is giving output 83?

Because, it's taken as octal base (8), since that numeral have 0 in leading. So, it's corresponding decimal value is 10.

012 :

(2 * 8 ^ 0) + (1 * 8 ^ 1) = 10

0123 :

(3 * 8 ^ 0) + (2 * 8 ^ 1) + (1 * 8 ^ 2) = 83

Upvotes: 25

user3463521
user3463521

Reputation: 578

You are assigning a constant to a variable using an octal representation of an type int constant. So the compiler gets the integer value out of the octal representation 010 by converting it to the decimal representation using this algorithm 0*8^0 + 1+8^1 = 10 and then assign j to 10. Remember when you see a constant starting with 0 it's an integer in octal representation. i.e. 0111 is not 1 hundred and 11 but it's 1*8^0 + 1*8^1 + 1*8^2.

Upvotes: 2

Mark
Mark

Reputation: 2822

The leading zero means the number is being interpreted as octal rather than decimal.

Upvotes: 6

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