我有这样的声明:
假设 byte 的位值为
x
00101011。结果是x>>2
什么?
我该如何编程,有人可以解释我在做什么吗?
首先,你不能在java中移动a byte
,你只能移动anint
或a long
。所以byte
将首先进行促销,例如
00101011
->00000000000000000000000000101011
或者
11010100
->11111111111111111111111111010100
现在,x >> N
意味着(如果您将其视为一串二进制数字):
00000000000000000000000000101011 >> 2
->00000000000000000000000000001010
11111111111111111111111111010100 >> 2
->11111111111111111111111111110101
二进制 32 位00101011
为
00000000 00000000 00000000 00101011
,结果是:
00000000 00000000 00000000 00101011 >> 2(times)
\\ \\
00000000 00000000 00000000 00001010
将 43 的位向右移动距离 2;用左侧的最高(符号)位填充。
结果为 00001010,十进制值为 10。
00001010
8+2 = 10
当您右移 2 位时,您会删除 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
这基本上与将 int 除以 2、2 次相同。
在 Java 中
byte b = (byte) 16;
b = b >> 2;
// prints 4
System.out.println(b);
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
>>
是算术右移运算符。第一个操作数中的所有位都移动了第二个操作数指示的位数。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。(这是为了使负数保持负数。)
这是您的具体情况:
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
如果您想查看您的数字的 bitString 表示,您可以使用例如此 API。罕见的数学
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
编辑:更改 api 链接并添加一个小示例
byte x = 51; //00101011
byte y = (byte) (x >> 2); //00001010 aka Base(10) 10
您不能像00101011
在 Java 中那样编写二进制文字,因此可以改为用十六进制编写:
byte x = 0x2b;
要计算你的结果,x >> 2
你可以准确地写出来并打印结果。
System.out.println(x >> 2);
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 1010