Java Algorithms-Symbols >>,>>>,<<

>> Right Shift

Move right, the Tao in binary, assuming that a 32-bit int represents a 64, then the high is 0, so when we move the entire binary number right, such as 0100000 >> 2 = 0001000, you can see the number of two bits after the right shift becomes 8, you can analyze In fact, the right shift is an operation divided by 2.

Example: For non-2,4,8,16,64 the number can also be tested:

SYSTEM.OUT.PRINTLN (3 >> 1); System.out.println (5 >> 1); System.out.println (>> 1);

The results were 1,2,31.

So the right shift is a process that divides the 2, the right one is except once, N is the n-th.

Also note that the >> is signed , that is, its high-level complement is determined by the highest bit, positive number of the highest bit is 0, negative number of the highest bit is 1, so negative >> or negative.

>>> Unsigned Right Shift

The principle is the same as above, and the difference is that its highs are always supplemented by the.

        SYSTEM.OUT.PRINTLN (3 >>> 1);        System.out.println (5 >>> 1);        System.out.println (>>> 1);        System.out.println (>>> 1);        System.out.println (-63 >>> 1);

The results were:

1232312147483616

You can see that the >>> symbol is the same as the >> symbol when calculating a positive number, and moves the position change if it is negative

Since 64 exists in the computer as: 11111111111111111111111111000000, then the unsigned right shift one becomes 01111111111111111111111111100000, which is equal to 2147483616

So do not use inertial thinking to think about these bit operations, and be sure to understand how the binary works.

<< left Shift

System.out.println ( -64 << 1);  System.out.println (<< 1);  System.out.println (<< 1);

The result is -128,128,50, apparently by 2 .

The left post-shift low is added by the

There is no <<< in Java, since the low post left is definitely added by the

Java Algorithms-Symbols >>,>>>,<<