按位运算符java

Question

my code is

final int CONST_1 = 1010;
final int CONST_2 = 1011;

System.out.println("CONST_1 & CONST_2: " + Integer.toBinaryString(CONST_1 & CONST_2));
System.out.println("CONST_1 ^ CONST_2: " + Integer.toBinaryString(CONST_1 ^ CONST_2));
System.out.println("CONST_1 | CONST_2: " + Integer.toBinaryString(CONST_1 | CONST_2));
System.out.println("~CONST_1 : " + Integer.toBinaryString(~CONST_1));

Output is

CONST_1 & CONST_2: 1111110010
CONST_1 ^ CONST_2: 1
CONST_1 | CONST_2: 1111110011
~CONST_1 : 11111111111111111111110000001101

In my opinion it's wrong and it should be:

CONST_1 & CONST_2: 1010
CONST_1 ^ CONST_2: 1
CONST_1 | CONST_2: 1011
~CONST_1 : 101

Please explain me why I have such result. Thanks!

Answer 1

Change this:

final int CONST_1 = 1010;
final int CONST_2 = 1011;

to this:

final int CONST_1 = 0b1010;
final int CONST_2 = 0b1011;

Don't forget that literals are decimal by default. You clearly wanted them to be binary.

---

Binary literals require Java 1.7. So if that's not available, you can go with this:

final int CONST_1 = Integer.parseInt("1010",2);
final int CONST_2 = Integer.parseInt("1011",2);

Answer 2

CONST_1 is 1010 in decimal. The binary value of CONST_1 is 1111110010 . Similarly CONST_2 is 1111110011 .

Does the result make more sense now?

Answer 3

I think you know what is meant by a literal. If not, please refer: Java Literals and Literal .

Now, Integer and Floating-Point literals are decimal by default in Java. So, the value 1010 you typed above will be decimal 1010 . ie., One thousand and ten . If you want them to be binary (it is clear from the question), there are many possibilities.

• Use equivalent decimal value.

You may use the decimal equivalent of the binary value you want to represent. Here, for example, decimal equivalent of binary 1010 is 10 , and that of binary 1011 is 11 .

final int CONST_1 = 10;
final int CONST_2 = 11;

• Use wrapper classes parse method

Each wrapper class has a parse method, which takes the base of number system also as argument. So

final int CONST_1 = Integer.parseInt("1010", 2);
final int CONST_2 = Integer.parseInt("1011", 2);

• If you are using Java 7, use binary literals

Binary literals are not supported in older versions of Java. Java 7 introduces binary literals. See features .

final int CONST_1 = 0b1010;
final int CONST_2 = 0b1011;

Answer 4

Don't confuse an integer composed of only 0 and 1 with the binary representation of the integer...

The integer 1010 is 1111110010 in binary, so the results are correct.

Answer 5

Your numbers aren't binary. They are written in decimal. You want to prepend a 0b to tell Java 7 this int is in binary. If you aren't using Java 7 there is no binary literal syntax so you can do this Integer.parseInt("1010", 2) as a work around or use HEX literal notation:

final int CONST_1_BINARY = 0b1010;
final int CONST_1_DECIMAL = 1010;

if( CONST_1_BINARY == CONST_1_DECIMAL ) {
   System.out.println("They are the same!");
} else {
   System.out.println("They are NOT the same.");
}