There are six basic bitwise operators: AND, OR, XOR, negation, shift left, and shift right:
Bit operation example operation
| Bit operation | Features | Example |
|---|---|---|
| x >> 1 | Remove the last digit | 101101->10110 |
| x << 1 | Add a 0 at the end | 101101->1011010 |
| x << 1 | Add a 1 at the end | 101101->1011011 |
| x|1 | Change the last digit to 1 | 101100->101101 |
| x & -2 | Change the last digit to 0 | 101101->101100 |
| x ^ 1 | The last bit is negated | 101101->101100 |
| x | (1 << (k-1)) | Turn the kth bit from the right to 1 | 101001->101101,k=3 |
| x & ~ (1 << (k-1)) | Turn the kth bit from the right to 0 | 101101->101001,k=3 |
| x ^(1 <<(k-1)) | The kth from the right is reversed | 101001->101101,k=3 |
| x & 7 | Take the last three | 1101101->101 |
| x & (1 << k-1) | Take the last k bits | 1101101->1101,k=5 |
| x >> (k-1) & 1 | Take the kth from the right | 1101101->1,k=4 |
| x | ((1 << k)-1) | Turn the last k bits into 1 | 101001->101111,k=4 |
| x ^ (1 << k-1) | Invert the last k bits | 101001->100110,k=4 |
| x & (x+1) | Turn the consecutive 1s on the right into 0 | 100101111->100100000 |
| x | (x+1) | Turn the first 0 from the right into 1 | 100101111->100111111 |
| x | (x-1) | Turn the consecutive 0s on the right into 1 | 11011000->11011111 |
| (x ^ (x+1)) >> 1 | Take the consecutive 1s on the right | 100101111->1111 |
| x & -x | Remove the left side of the first 1 from the right | 100101000->1000 |
| x&0x7F | Take the last 7 | 100101000->101000 |
| x& ~0x7F | Is it less than 127 | 001111111 & ~0x7F->0 |
| x & 1 | Parity | 00000111&1->1 |
Two considerations for using bit operations:
- Bit operations can only be used to reshape data. Bit operations on float and double types will be reported by the compiler.
- The operation priority of bit operators is relatively low, because try to use parentheses to ensure the order of operations
1. Determine whether a value is an integer power of 2
Problem-solving ideas:
There is only one 1 in the highest bit of the binary corresponding to the integer power of 2, such as: 8, the binary is 1000; 4, the binary is 0100,
Then subtract 1 from the number and perform the AND operation with the number. After subtracting 1, the binary is obtained: 7, the binary is 0111; 3, the binary is 0011, we can see that 8&7 is 0,
4&3 is 0
So, if n is an integer power of 2, then the result of n & (n-1) must be 0:
The value of n must be greater than 0
1. public class Main {
3. public static void main(String[] args) {
4. int n = 8;
5. if ((n & (n-1)) == 0){
6. System.out.println("整数的二次方 true");
7. }else{
8. System.out.println("不是整数的二次方");
9. }
10. }
11. }
2. Use bit operation to exchange two numbers [Do not use intermediate variables]
XOR
1. public class Main {
3. public static void main(String[] args) {
4. int n = 8, m = 10;
5. n ^= m;
6. m ^= n;
7. n ^= m;
8. System.out.println(n + ", " + m);
9. }
10. }
For example: a = 13, b = 6:
The binary of a is 13 = 8 + 4 + 1 = 1101 (binary)
The binary of b is 6 = 4 + 2 = 110 (binary)
- a ^= b a = 1101 ^ 110 = 1011;
- b ^ = ab = 110 ^ 1011 = 1101; Immediately b == 13
- a ^ = ba = 1011 ^ 1101 = 110; Immediately a == 6
Other methods, use addition and subtraction
1. public class Main {
3. public static void main(String[] args) {
4. int n = 8, m = 10;
5. n = n + m;
6. m = n - m;
7. n = n - m;
8. System.out.println(n + ", " + m);
9. }
10. }
3. Calculate how many 1s are in the binary representation of a 32-bit integer
Use x & (x-1) to eliminate the last digit 1, and calculate how many times it has been eliminated.
Such as:
13： 1101
12： 1100
Phase and: 1100, eliminate the last digit
1. public class Main {
3. public static void main(String[] args) {
4. // 计算在一个 32 位的整数的二进制表示中有多少个 1
5. int m = 13, num = 0;
6. while (true){
7. if (m == 0) break;
8. m &= (m-1);
9. num ++;
10. }
11. System.out.println(num);
12. }
14. }
4. Positive numbers become negative numbers, or negative numbers become positive numbers
The conversion sign only needs to be reversed and then add 1
1. public class Main {
3. public static void main(String[] args) {
4. // 计算在一个 32 位的整数的二进制表示中有多少个 1
5. int m = -13;
6. int changeM = ~m + 1;
7. System.out.println(changeM);
8. }
10. }
5. Determine the parity of a value
As long as it is decided according to whether the least bit is 0 or 1, 0 means even number, 1 means odd number, so only need to AND 1.
Therefore, if ((a & 1) == 0) can be used instead of if (a% 2 == 0) to determine whether a is even.
1. public class Main {
3. public static void main(String[] args) {
4. int m = -14;
6. if ((m & 1) == 1){
7. System.out.println("ji");
8. }else{
9. System.out.println("ou");
10. }
11. }
13. }
6. Multiply by 2 to the power of m operation
The operation of multiplying by 2, that is, 2 to the power of 1, and shifting 1 bit to the left
System.out.println(10<<1);Derive the extension:
Multiplied by 2 to the power of m
System.out.println(10<<2); // 乘以 2的2次方，相当于乘以 4 7. Divide by 2 operation (the operation of negative odd numbers is not available)
System.out.println(10>>1);8. Convert to absolute value
1. public class Main {
3. public static void main(String[] args) {
4. int n = 12;
6. System.out.println(0 >> 31); // 0
7. System.out.println(10 >> 31); // 0
8. System.out.println(-10 >> 31); // -1
10. System.out.println((n ^ (n >> 31)) - (n >> 31)); // 12
12. }
14. }
- First: n>>31 get the sign of n
If n is a positive number, n>>31 is equal to 0; if n is a negative number, n>>31 is equal to -1. If n is a positive number, n^0-0 is unchanged;
- If n is a negative number n^-1, you need to calculate the complement of n and -1, and then take the complement after the exclusive OR. As a result, the sign of n changes and the absolute value is subtracted by 1, and then minus -1 is the absolute value
9. Determine whether the signs of the two numbers are the same
true means x and y have the same sign, false means x and y have opposite signs.
System.out.println((a ^ b) > 0);10. Find the average of two integers (int)
System.out.println((a+b) >> 1);11. Find the maximum of two integers
1. int max(int a,int b){
2. return b & ((a-b) >> 31) | a & (~(a-b) >> 31);
3. /*如果a>=b,(a-b)>>31为0，否则为-1*/
4. }
12. Find the minimum of two integers
1. int min(int a,int b){
2. return a & ((a-b) >> 31) | b & (~(a-b) >> 31);
3. /*如果a>=b,(a-b)>>31为0，否则为-1*/
4. }
13. Addition of two integers
Use ^ and to & add two integers
- XOR of two numbers: It is equivalent to adding two numbers without considering carry;
- Two numbers are ANDed and shifted to the left by one position: equivalent to finding a carry;
13+11 = ？;
13 in binary 1 1 0 1 -----a 13
11 binary 1 0 1 1 -----b 11
(a&b) <<1 -> 1 0 0 1 0 -----d 18
a^b -> 0 1 1 0 -----e 6
(d&e) <<1 -> 0 0 1 0 0 ------f 4
d^e -> 1 0 1 0 0 -----g 20
(f&g) <<1 -> 0 1 0 0 0 ------h 8
f^g -> 1 0 0 0 0 ------i 16
(h&i) <<1 -> 0 0 0 0 0 ------h 0 ---- -------- There is no carry, then exit the loop
h^i -> 1 1 0 0 0 ------i 24
1. private static int getSum(int a, int b) {
2. if (a == 0) return b;
3. if (b == 0) return a;
4. while (b != 0) {
5. int carry = a & b; // 得到有进位的位置
6. a = a ^ b; // 直接相加，但是没有进位
7. b = carry << 1; // 得到进位
8. }
9. return a;
10. }
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