Bitwise operation 111&011=9, how?

Question

The bitwise operation 111&011 is giving output as 9 in java. how? 111&011 should be 011, ie 3.

Answer 1

You're specifying the numbers as decimal and octal , not as binary as you assume.

           number  | decimal value | binary  
           --------- -----------------------
(decimal)  111     | 111           | 1101111 
(octal)    011     | 9             | 0001001     

so:

   1101111
&     1001
   -------
   0001001

and 1001 = 2^3 + 1 = 9

to check it out use Integer.toBinaryString

public static void main (String[] args) {
  System.out.println(
    Integer.toBinaryString(111));

  System.out.println(
    Integer.toBinaryString(011));

  System.out.println(111 & 011);
}

code on ideone

Answer 2

Complementing yaitloutou's answer : an integer literal can be represented in different bases in Java:

• 0 , just the digit zero (base does not matter, but specified as decimal)
• decimal : a non-zero decimal digit eventually followed by decimal digits, eg 20
• hexadecimal : 0x followed by one or more hexadecimal digits, eg 0x14
• octal : 0 followed by one or more octal digits, eg 024
• binary : 0b followed by zeros and ones, eg 0b10100

The underscore _ can be used to separate digits, it will be ignored, eg 0b0001_0100

See the Java Language Specification 3.10.1 for more details.