How do shift operators work in Java? [duplicate]

I am trying to understand the shift operators and couldn't get much. When I tried to execute the below code

System.out.println(Integer.toBinaryString(2 << 11));
System.out.println(Integer.toBinaryString(2 << 22));
System.out.println(Integer.toBinaryString(2 << 33));
System.out.println(Integer.toBinaryString(2 << 44));
System.out.println(Integer.toBinaryString(2 << 55));

I get the below

1000000000000    
100000000000000000000000    
100    
10000000000000    
1000000000000000000000000    

Could somebody please explain?

Solution 1:

System.out.println(Integer.toBinaryString(2 << 11)); 

Shifts binary 2(10) by 11 times to the left. Hence: 1000000000000

System.out.println(Integer.toBinaryString(2 << 22)); 

Shifts binary 2(10) by 22 times to the left. Hence : 100000000000000000000000

System.out.println(Integer.toBinaryString(2 << 33)); 

Now, int is of 4 bytes,hence 32 bits. So when you do shift by 33, it's equivalent to shift by 1. Hence : 100

Solution 2:

2 from decimal numbering system in binary is as follows

10

now if you do

2 << 11

it would be , 11 zeros would be padded on the right side

1000000000000

The signed left shift operator "<<" shifts a bit pattern to the left, and the signed right shift operator ">>" shifts a bit pattern to the right. The bit pattern is given by the left-hand operand, and the number of positions to shift by the right-hand operand. The unsigned right shift operator ">>>" shifts a zero into the leftmost position, while the leftmost position after ">>" depends on sign extension [..]

left shifting results in multiplication by 2 (*2) in terms or arithmetic

For example

2 in binary 10, if you do <<1 that would be 100 which is 4

4 in binary 100, if you do <<1 that would be 1000 which is 8

Also See

• absolute-beginners-guide-to-bit-shifting