We know that all numbers in the program are stored in binary form in computer memory, and bit operations are to directly operate on the binary bits of integers in memory. For example, the and operation is originally a logical operator, but the AND operation can also be performed between integers.
Bit operations mainly include shift operations and logical operations. Next we will talk about shift operations and logical operations respectively.
Shift operation:
Left shift: The operator is <<, move to the left, add 0 to the low bit on the right, discard the high bit on the left, treat binary as an integer, shift left by 1 bit is equivalent to multiplying by 2. Unsigned right shift: The operator is >>>, moving to the right, discarding the right side, and adding 0 to the left side. Signed right shift: The operator is >>, move to the right, and discard the right side. The value added to the left depends on the original highest bit. If it is originally 1, add 1, and if it is 0, add 0. Think of binary as For an integer, shifting it right by 1 bit is equivalent to dividing by 2.
For example:
int a = 4; // 100 a = a >> 2; // 001,等于1 a = a << 3 // 1000,变为8
Logical operations include:
Bitwise AND&: both bits are 1 to be 1
Bitwise OR|: As long as one bit is 1, it is 1
int a = ...; a = a & 0x1 // 返回0或1,就是a最右边一位的值。 a = a | 0x1 //不管a原来最右边一位是什么,都将设为1
int i = 1;// 二进制存储方式为00000000000000000000000000000001 int j = 5;// 二进制存储方式为00000000000000000000000000000101 int k = 6;// 二进制存储方式为00000000000000000000000000000110 if ((i & j) == 1) { System.out.println("j的最低位为1,为奇数"); } if ((i & k) == 0) { System.out.println("k的最低位为0,为偶数"); }
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)Scenario 3: Simple collection processingNo nonsense, just look at the code:
(n&(n-1))==0
public class SimpleSet { public static final int A = 0x01;// 最后四位为0001 public static final int B = 0x02;// 最后四位为0010 public static final int C = 0x04;// 最后四位为0100 public static final int D = 0x08;// 最后四位为1000 private int set = 0x00;// 初始0000,空集合 public void add(int i) {// 将i对应位的值置为1,重复add不影响。默认传入值为ABCD之一,此处省去边界判断 set |= i; } public boolean contain(int i) {// 判断相应位置是否为1 return (set & i) == i; } public boolean remove(int i) {// 来不及不解释了快看代码 if (contain(i)) { set -= i; return true; } else { return false; } } }
public static void main(String[] args) { SimpleSet set = new SimpleSet(); System.out.println(set.contain(A)); set.add(B); System.out.println(set.contain(A)); System.out.println(set.contain(B)); set.add(A); set.add(C); System.out.println(set.contain(A)); set.remove(A); System.out.println(set.contain(A)); System.out.println(set.remove(A)); System.out.println(set.contain(C)); }
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