"Original" Java Shift operations

To learn the shift operation, you first need to know the number of bits involved in the shift operation, so let's review the placeholder for the Java base type.

• Java base type

Summary List of Java Foundation types

Type	Number of binary digits	Maximum Value	Minimum value	Initialize value	Representation	with symbols
Char	8	127 (2^7-1)	-128 ( -2^7)	0	Two-step complement	Is
Short	16	32767 (2^15-1)	-32768 ( -2^15)	0	Two-step complement	Is
Int	32	2,147,485,647 (2^31-1)	-2,147,483,648 ( -2^31)	0	Two-step complement
Long	64	9,223,372,036,854,775,807 (2^63-1)	-9,223,372,036,854,775,808 ( -2^63)	0L	Two-step complement	Is
Float	32	3.4028235E38	1.4E-45	0.0f	IEEE 754 standard floating-point number	Is
Double	64	1.7976931348623157E308	4.9E-324	0.0f	IEEE 754 standard floating-point number	Is
Boolean	Pending analysis	Only true and False	Only true and False	False	The JVM uses 0/1 to indicate

The difference between float and double is that one is a single-precision floating-point number, and the other is a double-precision floating-point number.

A floating-point number is a representation of an exponential type within a machine, which can be decomposed into four parts: a symbol, a mantissa, an exponential symbol, an exponent

Both the number sign bit and the exponent sign bit are one, indicating positive or negative.

For float, the exponent portion occupies 8 bits, wherein the exponent symbol is one digit, the value part is 7 bits, the tail part occupies 24 bits; for double, the exponent portion is 16 bits, the exponent part is 15 bits, and the tail part occupies 48 bits. So double is larger and more accurate than the number represented by float, but at the same time it is twice times the memory consumption.

The Boolean majority problem depends on the situation, in the JVM, the operation of the Boolean value is replaced by the int type, so it occupies 32 bits; If you define a byte array of type Boolean, the JVM compiles it to a byte array type, which takes up 8 bits. In fact, here also listen to doubts, why a Boolean can occupy 8, one will occupy 32? Personal understanding of the feeling is that when the real Boolean, should be only 1 bits, not 0 is 1, but the specific compile time other occupancy may be used for other purposes.

OK, next, let's look at the problem with the shift operation.

the shift operators for Java include  three kinds:, >> (with symbol right shift) and >>> (unsigned Right shift).

• left shift operator <<

Usage: value << number

Principle: Discard the highest bit, low 0

Note: The Char,byte and short type are shifted to the int type before the shift operation.

When the number of bits moved exceeds the number of digits of the number of moves, a modulo operation, such as 45<<34, is moved 2 bits, because 34%32=2.

Instance:

1. Example procedures

 Public class Test {    publicstaticvoid  main (String args[]) {          int num =0x40000000;           SYSTEM.OUT.PRINTLN (num);          = num << 1;         SYSTEM.OUT.PRINTLN (num);    }}

2. Program Results

1073741824-2147483648

A typical use case is selected, which shows that the symbol bit is also moved together, the number bit is 1 and moves to the sign bit, which turns the numbers into negative numbers.

• Right shift with symbols >>

Usage: value >> number

Principle: The symbol bit is unchanged, the left side of the symbol is added, and at the same time the lowest bit to discard

Note: The Char,byte and short type are shifted to the int type before the shift operation.

When the number of bits moved exceeds the number of digits of the number of moves, a modulo operation, such as 45<<34, is moved 2 bits, because 34%32=2.

Instance:

1. Example procedures

 Public class Test {    publicstaticvoid  main (String args[]) {          int NUM1 =0x80000000;            int num2 = 0x00000003;         System.out.println (NUM1);         System.out.println (num2);          = Num1 >> 2;          = num2 >> 1;         System.out.println (NUM1);         System.out.println (num2);    }}

2. Running Results

-21474836483-5368709121

NUM1 shows that the symbol moves to the right, with symbol movement.

Num2 shows that the lows are discarded and changed from 3 to 1.

• Unsigned Right Shift >>>

Usage: value >>> number

Principle: The symbol is unchanged, the left side is 0, and at the same time the lowest bit to discard

Note: The Char,byte and short type are shifted to the int type before the shift operation.

When the number of bits moved exceeds the number of digits of the number of moves, a modulo operation, such as 45<<34, is moved 2 bits, because 34%32=2.

Instance:

1. Example procedures

 Public class Test {    publicstaticvoid  main (String args[]) {          int NUM1 =0x80000000;            int num2 = 0x00000003;         System.out.println (NUM1);         System.out.println (num2);          = NUM1 >>> 1;          = num2 >>>;         System.out.println (NUM1);         System.out.println (num2);    }}

2. Running Results

-2147483648310737418241

As you can see from NUM1, the sign bit also moves along with the digital digits.

"Original" Java Shift operations