我有以下陈述:
假设字节
我该如何编程并有人能解释一下它是做什么的?
假设字节
x
的位值为00101011。那么x>>2
的结果是什么?我该如何编程并有人能解释一下它是做什么的?
x
的位值为00101011。那么x>>2
的结果是什么?00101011
-> 00000000000000000000000000101011
或者
11010100
-> 11111111111111111111111111010100
现在,x >> N
的意思是(如果你将其视为一串二进制数字):
00000000000000000000000000101011 >> 2
-> 00000000000000000000000000001010
11111111111111111111111111010100 >> 2
-> 11111111111111111111111111110101
二进制数00101011
的32位表示为:
00000000 00000000 00000000 00101011
,计算结果如下:
00000000 00000000 00000000 00101011 >> 2(times)
\\ \\
00000000 00000000 00000000 00001010
将43的二进制位向右移动2位;左侧填充最高(符号)位。
结果为00001010,十进制值为10。
00001010
8+2 = 10
当你向右移动2位时,你会丢掉最低的2位。因此:
x = 00101011
x >> 2
// now (notice the 2 new 0's on the left of the byte)
x = 00001010
这本质上等同于将 int 值除以 2,两次。
在 Java 中
byte b = (byte) 16;
b = b >> 2;
// prints 4
System.out.println(b);
incompatible types: possible lossy conversion from int to byte
。 - Zhou Haibo这些示例涵盖了应用于正数和负数的三种类型的移位:
// 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
>>
是算术右移操作符。第一个操作数中的所有位将向右移动由第二个操作数指定的位数。结果中最左边的位被设置为与原始数字中最左边的位相同的值。(这样负数仍然保持负数。)
以下是您的具体情况:
00101011
001010 <-- Shifted twice to the right (rightmost bits dropped)
00001010 <-- Leftmost bits filled with 0s (to match leftmost bit in original number)
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
在Java中,您无法像00101011
那样编写二进制字面量,因此您可以改为以十六进制编写:
byte x = 0x2b;
x >> 2
的结果,你只需直接写出这个表达式并打印出结果即可。System.out.println(x >> 2);
0b00101011
。 - Alan Kruegerbyte x = 51; //00101011
byte y = (byte) (x >> 2); //00001010 aka Base(10) 10
bitString = org.uncommons.maths.binary.BitString.new(java.math.BigInteger.new("12").toString(2))
bitString.setBit(1, true)
bitString.toNumber => 14
编辑:更改了 API 链接并添加了一个小例子
00101011 = 43 十进制
class test {
public static void main(String[] args){
int a= 43;
String b= Integer.toBinaryString(a >> 2);
System.out.println(b);
}
}
输出:
101011 变成了 1010