bithifting如何在Java中工作？

Question

我有这样的声明：

  

假设字节x的位值为00101011. x>>2的结果是什么？

我如何编程并且有人可以解释我在做什么？

Answer 1

首先，你可以不在java中移动byte，你只能移动int或long。因此，byte将首先进行促销，例如

00101011 - ＆gt; 00000000000000000000000000101011

或

11010100 - ＆gt; 11111111111111111111111111010100

现在，x >> N表示（如果您将其视为二进制数字字符串）：

• 最右边的N位被丢弃
• 根据需要复制最左边的位以将结果填充为原始大小（32或64位），例如

00000000000000000000000000101011 >> 2 - ＆gt; 00000000000000000000000000001010

11111111111111111111111111010100 >> 2 - ＆gt; 11111111111111111111111111110101

Answer 2

00101011的二进制32位是

00000000 00000000 00000000 00101011，结果是：

  00000000 00000000 00000000 00101011   >> 2(times)
 \\                                 \\
  00000000 00000000 00000000 00001010

将位43向右移动距离2;填充左侧最高（符号）位。

结果为00001010，十进制值为10。

00001010
    8+2 = 10

Answer 3

右移2位时，丢弃2个最低有效位。所以：

x = 00101011

x >> 2

// now (notice the 2 new 0's on the left of the byte)
x = 00001010

这与将int分为2次2基本相同。

在Java中

byte b = (byte) 16;
b = b >> 2;
// prints 4
System.out.println(b);

Answer 4

这些例子涵盖了应用于正数和负数的三种类型的转变：

// Signed left shift on 626348975
00100101010101010101001110101111 is   626348975
01001010101010101010011101011110 is  1252697950 after << 1
10010101010101010100111010111100 is -1789571396 after << 2
00101010101010101001110101111000 is   715824504 after << 3

// Signed left shift on -552270512
11011111000101010000010101010000 is  -552270512
10111110001010100000101010100000 is -1104541024 after << 1
01111100010101000001010101000000 is  2085885248 after << 2
11111000101010000010101010000000 is  -123196800 after << 3


// Signed right shift on 626348975
00100101010101010101001110101111 is   626348975
00010010101010101010100111010111 is   313174487 after >> 1
00001001010101010101010011101011 is   156587243 after >> 2
00000100101010101010101001110101 is    78293621 after >> 3

// Signed right shift on -552270512
11011111000101010000010101010000 is  -552270512
11101111100010101000001010101000 is  -276135256 after >> 1
11110111110001010100000101010100 is  -138067628 after >> 2
11111011111000101010000010101010 is   -69033814 after >> 3


// Unsigned right shift on 626348975
00100101010101010101001110101111 is   626348975
00010010101010101010100111010111 is   313174487 after >>> 1
00001001010101010101010011101011 is   156587243 after >>> 2
00000100101010101010101001110101 is    78293621 after >>> 3

// Unsigned right shift on -552270512
11011111000101010000010101010000 is  -552270512
01101111100010101000001010101000 is  1871348392 after >>> 1
00110111110001010100000101010100 is   935674196 after >>> 2
00011011111000101010000010101010 is   467837098 after >>> 3

Answer 5

>>是算术右移运算符。第一个操作数中的所有位都移位了第二个操作数指示的位数。结果中最左边的位设置为与原始数字中最左边的位相同的值。 （这是负数仍为负数。）

以下是您的具体案例：

00101011
  001010 <-- Shifted twice to the right (rightmost bits dropped)
00001010 <-- Leftmost bits filled with 0s (to match leftmost bit in original number)

Answer 6

public class Shift {
 public static void main(String[] args) {
  Byte b = Byte.parseByte("00101011",2);
  System.out.println(b);
  byte val = b.byteValue();
  Byte shifted = new Byte((byte) (val >> 2));
  System.out.println(shifted);

  // often overloked  are the methods of Integer

  int i = Integer.parseInt("00101011",2);
  System.out.println( Integer.toBinaryString(i));
  i >>= 2;
  System.out.println( Integer.toBinaryString(i));
 }
}

输出：

43
10
101011
1010

Answer 7

你不能在Java中编写像00101011这样的二进制文字，所以你可以用十六进制编写它：

byte x = 0x2b;

要计算x >> 2的结果，您可以准确地写出并打印结果。

System.out.println(x >> 2);

Answer 8

byte x = 51; //00101011
byte y = (byte) (x >> 2); //00001010 aka Base(10) 10

Answer 9

您可以使用例如如果您想查看数字的bitString表示，请使用此API。 Uncommons Math

示例（在jruby中）

bitString = org.uncommons.maths.binary.BitString.new(java.math.BigInteger.new("12").toString(2))
bitString.setBit(1, true)
bitString.toNumber => 14

修改：更改了api链接并添加了一些示例

Answer 10

00101011 =十进制43

class test {    
    public static void main(String[] args){
       int a= 43;       
       String b= Integer.toBinaryString(a >> 2);        
       System.out.println(b);
    }   
}

输出：

101011变为1010