Java位运算与运算的用途

落日映苍穹つ 2024-03-26 23:17 10阅读 0赞

#### 目录标题 ####

*  ① &0清零
 *  ② &-1值不变
 *  ③ 根据 ① + ②
 *  ④ 判断奇偶
 *  ⑤ 判断一个整数是否为2的整数次方
 *  ⑥ 【useless】不用 for 循环找到二进制最右侧的1 → \`num & (-num)\`
 *  ⑦ 【useless】将二进制表示的最右侧的1置0 → \`num & (-num) ^ num\`

## ① &0清零 ##

> 一个数与一个各位都为0的数值相与，结果为零

举例：
       0000 0101 1010
    &  0000 0000 0000
    ---------------------
       0000 0000 0000

int[] arr = {
        1, 2, 3, 4, 5, 6, 8};
    for (int i = 0; i < arr.length; i++) {
        
        int res = arr[i] & 0;
        System.out.print(res);
    }
    // 0000000

## ② &-1值不变 ##

举例：
       0000 0101 1010
    &  1111 1111 1111 
    ---------------------
       0000 0101 1010

## ③ 根据 ① + ② ##

获取某个二进制数的指定位 → `10101110 & 00001111 = 00001110`  
把某个二进制数的指定位置为0 → `10101110 & 00001111 = 00001110`

举例：
       0000 0101 1010
    &  0000 1111 0000
    ---------------------
       0000 0101 0000

## ④ 判断奇偶 ##

二进制未位为0就是偶数，为1就是奇数
    
    Integer.toBinaryString(0)  0000
    Integer.toBinaryString(1)  0001
    Integer.toBinaryString(2)  0010
    Integer.toBinaryString(3)  0011
    Integer.toBinaryString(4)  0100

# 判断奇数
    if ((a & 1) == 1)
    
    # 判断偶数
    if ((a & 1) == 0)
    if (a % 2 == 0)
    
    System.out.println((5 & 1) == 1 ? "奇数" : "偶数"); // 奇数
    System.out.println((6 & 1) == 1 ? "奇数" : "偶数"); // 偶数
    System.out.println((0 & 1) == 1 ? "奇数" : "偶数"); // 偶数
    System.out.println((-1 & 1) == 1 ? "奇数" : "偶数"); // 奇数
    System.out.println((-2 & 1) == 1 ? "奇数" : "偶数"); // 偶数

## ⑤ 判断一个整数是否为2的整数次方 ##

# 记住2的n次方法的二进制的样子
    (proFillZero(Integer.toBinaryString(0)));	    00000000000000000000000000000000
    (proFillZero(Integer.toBinaryString(2)));	    00000000000000000000000000000010
    (proFillZero(Integer.toBinaryString(4)));	    00000000000000000000000000000100
    (proFillZero(Integer.toBinaryString(8)));	    00000000000000000000000000001000
    (proFillZero(Integer.toBinaryString(16)));	    00000000000000000000000000010000
    (proFillZero(Integer.toBinaryString(32)));	    00000000000000000000000000100000
    (proFillZero(Integer.toBinaryString(64)));	    00000000000000000000000001000000
    (proFillZero(Integer.toBinaryString(128)));	    00000000000000000000000010000000
    (proFillZero(Integer.toBinaryString(256)));	    00000000000000000000000100000000
    (proFillZero(Integer.toBinaryString(512)));	    00000000000000000000001000000000
    (proFillZero(Integer.toBinaryString(1024)));	00000000000000000000010000000000
    (proFillZero(Integer.toBinaryString(2048)));	00000000000000000000100000000000
    (proFillZero(Integer.toBinaryString(4096)));	00000000000000000001000000000000
    (proFillZero(Integer.toBinaryString(8192)));	00000000000000000010000000000000

(proFillZero(Integer.toBinaryString(-2)));	    11111111111111111111111111111110
    (proFillZero(Integer.toBinaryString(-4)));	    11111111111111111111111111111100
    (proFillZero(Integer.toBinaryString(-8)));	    11111111111111111111111111111000
    (proFillZero(Integer.toBinaryString(-16)));	    11111111111111111111111111110000
    (proFillZero(Integer.toBinaryString(-32)));	    11111111111111111111111111100000
    (proFillZero(Integer.toBinaryString(-64)));	    11111111111111111111111111000000
    (proFillZero(Integer.toBinaryString(-128)));	11111111111111111111111110000000
    (proFillZero(Integer.toBinaryString(-256)));	11111111111111111111111100000000
    (proFillZero(Integer.toBinaryString(-512)));	11111111111111111111111000000000
    (proFillZero(Integer.toBinaryString(-1024)));	11111111111111111111110000000000
    (proFillZero(Integer.toBinaryString(-2048)));	11111111111111111111100000000000
    (proFillZero(Integer.toBinaryString(-4096)));	11111111111111111111000000000000
    (proFillZero(Integer.toBinaryString(-8192)));	11111111111111111110000000000000

(proFillZero(Integer.toBinaryString(7)));		00000000000000000000000000000111
    (proFillZero(Integer.toBinaryString(7 - 1)));	00000000000000000000000000000110
    
    (proFillZero(Integer.toBinaryString(4)));		00000000000000000000000000000100
    (proFillZero(Integer.toBinaryString(4 - 1)));	00000000000000000000000000000011
    
    这里减1会导致退1位，所以再和原来的取&，肯定为0, 即(num & (num-1)) == 0

private static void findLoca(int num) {
        
        boolean noSame;
        if (num >= 0) {
        
            noSame = (num & (num - 1)) == 0;
        } else {
        
            # 正数这个算法才好用，所以负数变成正数
            noSame = (-num & (-num - 1)) == 0;
        }
        System.out.println(noSame);
    }

## ⑥ 【useless】不用 for 循环找到二进制最右侧的1 → `num & (-num)` ##

已知 `num` & `num取反` = `0`，`-num` = `num取反+1`，  
故`num` & `-num` → `num` & `num取反+1` → 最后一个不同的`1`

private static void findLoca1(int num) {
        
        System.out.println(" 2进制   -1         : " + Integer.toBinaryString(-1));
        System.out.println("10进制  " + num + "       数: " + num);
        System.out.println(" 2进制  " + num + "    绝对值: " + proFillZero(Integer.toBinaryString(num)));
        System.out.println(" 2进制  " + num + " 绝对值取反: " + Integer.toBinaryString(-num - 1));
        System.out.println(" 2进制 " + (-num) + "      补码: " + Integer.toBinaryString(-num));
        int location = (num & (-num));
        // -num就是num取反+1得到的,num&num取反=0,num&-num→最后一个不同的1
        System.out.println(proFillZero(Integer.toBinaryString(location)));
    }
    
     2进制   -1         : 11111111111111111111111111111111
    10进制  500       数: 500
     2进制  500    绝对值: 00000000000000000000000111110100
     2进制  500 绝对值取反: 11111111111111111111111000001011
     2进制 -500      补码: 11111111111111111111111000001100
    00000000000000000000000000000100

## ⑦ 【useless】将二进制表示的最右侧的1置0 → `num & (-num) ^ num` ##

System.out.println("10进制  " + num + "       数: " + num);
    System.out.println(" 2进制  " + num + "    绝对值: " + proFillZero(Integer.toBinaryString(num)));
    System.out.println(" 2进制  " + num + " 绝对值取反: " + Integer.toBinaryString(-num - 1));
    System.out.println(" 2进制 " + (-num) + "      补码: " + Integer.toBinaryString(-num));
    
    int location = (num & (-num));
    // -num就是num取反+1得到的,num&num取反=0,num&-num→最后一个不同的1
    System.out.println(proFillZero(Integer.toBinaryString(location)));
    
    int location1 = num & (-num) ^ num;
    System.out.println(proFillZero(Integer.toBinaryString(location1)));
    
    10进制  500       数: 500
     2进制  500    绝对值: 000000000000000
     2进制  500 绝对值取反: 1111111111111111
     2进制 -500      补码: 11111111111111
    00000000000000000000000000000100
    00000000000000000000000111110000