将 int 左移时会发生什么

Question

我正在尝试理解为创建 my_hashset 而创建的这段代码的一部分：https://leetcode.com/problems/design-hashset/discuss/548792/Java-solution-faster-than-99

   class HashCode_from_leet {
    int[] bitArr;
    private static final int MAX = 100000;
    private static final int INT_SIZE = Integer.SIZE;

    public HashCode_from_leet() {
        bitArr = new int[MAX / INT_SIZE + 1];
    }
    
    public void add(int key) {  //pass in 5
        int ind = key / INT_SIZE;  
        int bit = key % INT_SIZE;
        int bitMask = 1 << bit;  //QUESTION: why does bitMask = 32?
        bitArr[ind] |= bitMask;  // QUESTION: what does '|=' mean?
    }
    
    // NOTE I REMOVED SOME METHODS 
    
    public boolean contains(int key) {
        int ind = key / INT_SIZE;
        int bit = key % INT_SIZE;
        int bitMask = 1 << bit;
        return (bitArr[ind] & bitMask) != 0;
    }
    
    
    public static void main(String[] args) {
        HashCode_from_leet hfl = new HashCode_from_leet();
        hfl.add(5);
        System.out.println(hfl.contains(5));
    }
}

当我将 5 传入 add(int key) 方法时 bitMask = 32

我不确定为什么 -- 我理解 << 表示向左移动，所以我们取 5 并将其向左移动，但会认为等于 5*10^2 吗？

Answer 1

当你 shift left X 的整数时，例如：Z =(Y << X) 它会实际执行此操作 : Z = (Y * (2^X))

当你 shift right X 的整数时，例如：Z =(Y >> X) 它会实际执行此操作 : Z = (Y / (2^X))

Answer 2

what happens when you shift an int left

顾名思义，将数字左移 n 位，乘以 2^n。下面的代码将帮助您更好地理解它。

public class Main {
    public static void main(String[] args) {
        int x = 5;
        System.out.println("In decimal: " + x + ", In Binary: "
                + "0000000000000000".substring(Integer.toBinaryString(x).length()) + Integer.toBinaryString(x));
        int y = x << 1;
        System.out.println("In decimal: " + y + ", In Binary: "
                + "0000000000000000".substring(Integer.toBinaryString(y).length()) + Integer.toBinaryString(y));
        int z = x << 2;
        System.out.println("In decimal: " + z + ", In Binary: "
                + "0000000000000000".substring(Integer.toBinaryString(z).length()) + Integer.toBinaryString(z));
    }
}

输出：

In decimal: 5, In Binary: 0000000000000101
In decimal: 10, In Binary: 0000000000001010
In decimal: 20, In Binary: 0000000000010100

Answer 3

通知行int bit = key % INT_SIZE; // which is 5 % 32 according to your code

所以现在bit = 5 // because 5 % 32 = 5

现在int bitMask = 1 << bit; // you're saying shit 1 to five bits to left

in 32 bits 1 = 000000000000000000000000000000000001

左移5位后变为：

000000000000000000000000000000000100000 = 32 //in binary that is why bitmask continas 32

现在我们来谈谈|=运算符

| 是一个按位或运算符，您只是将其用作复合赋值

bitArr[ind] |= bitMask is similar to bitArr[ind] = bitArr[ind] | bitmask

例如说bitArr[ind] equals 5 and bitmask equals 3

现在bitArr[ind] | bitmask意味着5 | 3

它将与二进制表示进行按位或运算

表示 101 | 011 变成 111 即 7

将 int 左移时会发生什么

What happens when you shift an int left

java

bit-shift

hashset