具有前导零输出的长变量

Question

I am not able to understand the why such output when long variable with leading zero .

public class Test{
        public static void main(String[] args) {
             long var1=00123l;
             long var2=123l;
             System.out.println("Variable 1--->"+var1);
             System.out.println("Variable 2--->"+var2);
             System.out.println(var1==var2);
        }
    }

output:

Variable 1--->83
Variable 2--->123
false

Answer 1

when a literal is prefixed with 0; java treats it as a octal number. When you print that same number, by default it prints in base(10) format. Hence 00123l is printed as 83 .

Answer 2

The leading zero turns 00123l into an octal literal, and 123 ₈ =83 ₁₀ .

From the JLS :

An octal numeral consists of an ASCII digit 0 followed by one or more of the ASCII digits 0 through 7 interspersed with underscores, and can represent a positive, zero, or negative integer.

When you print the value, it gets printed in base-10, so you see 83 .

Answer 3

When you add a leading zero to a value, it is interpreted as an octal value.

Integers can not store any leading zeros. When you need those, store the number as a String.

Answer 4

When you write an integer literal with leading zeroes, it is interpreted as an octal number. 00123 in octal is 83 in decimal.

Answer 5

00123 or 0123 or 0123l all are same for Java. You can test. They are octal numbers. If you convert them, you will find (1X8^2)+(1X8^1)(1X8^0)=64+16+3=83 . If you want you can store them as String.