Left Right Shift in Java

VERY IMPORTANCE SOURCES:
http://www.xyzws.com/javafaq/how-do-the-bitwise-shift-operators-work/12
http://stackoverflow.com/questions/10910913/how-do-shift-operators-work-in-java

Summary:
1. RIGHT SHIFT (>>)
In >>, the void bits are filled in with previous contents of the leftmost bit.(FOR BOTH POSITIVE AND NEGATIVE NUMBERS)
Note that right shifting ( >> ) always preserves the sign of the original number i.e. to say that a negative number will
stay negative while a positive number will stay positive after a right shift ( >> ).

Example:
Consider 12 whose binary equivalent is 00000000 00000000 00000000 00001100 (binary for 32 bits). 
Remember that the type operands (such as byte and short) are implicitly promoted to the int type before 
any shift operators are applied. Let's shift it right 2 times.

00000000 00000000 00000000 00001100 >> 2(times)
When shifting right ( >> ) the leftmost bits exposed by the right shift are filled in with previous contents
of the leftmost bit. This is the result of (12>>2):

00000000 00000000 00000000 00000011 (3)
Consider -12 whose binary equivalent is 11111111 11111111 11111111 11110100 (binary for 32 bits).
Let's shift it right 2 times.

11111111 11111111 11111111 11110100 >> 2(times)
When shifting right ( >> ) the leftmost bits exposed by the right shift are filled in with previous contents of 
the leftmost bit. This is the result of (-12>>2):

11111111 11111111 11111111 11111101 (-3)

2. Unsigned Right Shift Operator (>>>)
In >>, the void bits are filled with 0's.(FOR BOTH POSITIVE AND NEGATIVE NUMBERS)
Let's do 00000000 00000000 00000000 00001100 >>> 2 (12>>>2) and the result is 00000000 00000000 00000000 00000011 (3).
Let's do 11111111 11111111 11111111 11110100 >>> 2 (-12>>>2) and the result is 00111111 11111111 11111111 11111101 (1073741821).

3. LEFT SHIFT (<<)
In <<, the void left behind by the shift is filled by zero's. (FOR BOTH POSITIVE AND NEGATIVE NUMBERS)

Example: 
Consider 12 whose binary equivalent is 00000000 00000000 00000000 00001100 (binary for 32 bits). 
Remember that the type operands (such as byte and short) are implicitly promoted to the int type before any shift
operators are applied. Let's shift it left 2 times.

00000000 00000000 00000000 00001100 << 2(times)
When you shift left ( << ) the void left behind by the shift is filled by zero's. This is the result of (12<<2):

00000000 00000000 00000000 00110000 (48)
Consider -12 whose binary equivalent is 11111111 11111111 11111111 11110100 (binary for 32 bits). Let's shift it left 2 times.

11111111 11111111 11111111 11110100 << 2(times)
When you shift left ( << ) the void left behind by the shift is filled by zero's. This is the result of (-12<<2):

11111111 11111111 11111111 11010000 (-48)
Let's do -12<<28 and the result is

01000000 00000000 00000000 00000000 (1073741824)