20 bit operations programmers should know

While talking about bit operation, programmers usually think about its efficiency. In embedded programming and system core optimization, appropriate use of bit operations is always a fascinating. When finding a job, it will help you a lot if you use bit operations when writing codes. Mastering simple bit arithmetic skills is necessary.

1. Get maximum int value

1. int getMaxInt(){  
2.         return (1 << 31) - 1;//2147483647，  
   
3. } 
   

Another way:

1. int getMaxInt(){  
2.     return ~(1 << 31);//2147483647  
3. } 
   

One more way:

1. int getMaxInt(){
2.     return (1 << -1) - 1;//2147483647  
3. } 
   

In C, if you don't know how many bytes an int is.

1. int getMaxInt(){  
2.     return ((unsigned int) - 1) >> 1;//2147483647  
3. } 
   

2. Get minimum int value

1. int getMinInt(){  
2.     return 1 << 31;//-2147483648  
3.  } 
   

Another way

1. int getMinInt(){
   
2.     return 1 << -1;//-2147483648  
3. } 
   

3. Get maximum long value

C version:

1. long getMaxLong(){  
2.     return ((unsigned long) - 1) >> 1;//2147483647  
3. }
   

Java version :

1. long getMaxLong(){  
2.     return ((long)1 << 127) - 1;//9223372036854775807  
3. } 
   

4. Multiply 2

1. int mulTwo(int n){//计算n*2   
2.     return n << 1;  
3. } 
   

5. Divide by 2

1. int divTwo(int n){
   
2.     return n >> 1;//除以2  
3. } 
   

6. Multiple 2m

1. int mulTwoPower(int n,int m){
2.     return n << m;  
3. } 
   

7. Divide by 2^m

1. int divTwoPower(int n,int m){
2.     return n >> m;  
3. } 
   

8. Check whether a number is odd or even

1. boolean isOddNumber(int n){  
2.     return (n & 1) == 1;  
3. } 
   

9. Exchange values of two variables

1. void swap(int *a,int *b){     
2.     (*a) ^= (*b) ^= (*a) ^= (*b);     
3. } 
   

Common version

1. a ^= b;  
2. b ^= a;  
3. a ^= b; 
   

10. Get absolute value

1. int abs(int n){  
2. return (n ^ (n >> 31)) - (n >> 31);
   
3. } 
   

n>>31 will get the sign of n, if n is positive, then n>>31 is 0, if n is negative, n>>31 is -1. If n is positive, then n^0=n. If n is negative, then to calculate n^-1 , we need to calculate n and -1's complement and then do exclusive or operation. Then n will change the sign and the value is absolute value of n minus 1. So at last, do a minus -1 to get the absolute value.

11. Get bigger value of two values

1. int max(int a,int b){  
2.     return b & ((a-b) >> 31) | a & (~(a-b) >> 31);  
3. } 
   

C version:

1. int max(int x,int y){  
2.     return x ^ ((x ^ y) & -(x < y));  
3. } 
   

12. Get smaller value of two values

1. int min(int a,int b){  
2.     return a & ((a-b) >> 31) | b & (~(a-b) >> 31);  
   
3. } 
   

C version:

1. int min(int x,int y){  
2.     return y ^ ((x ^ y) & -(x < y));  
3. } 
   

13. Check whether two values have the same sign

1. boolean isSameSign(int x, int y){
2.     return (x ^ y) >= 0;
   
3. } 
   

14. Calculate 2ⁿ

1. int getFactorialofTwo(int n){//n > 0  
2.     return 2 << (n-1);
3. } 
   

15. Check whether a value can be divided by 2

1. boolean isFactorialofTwo(int n){  
2.     return n > 0 ? (n & (n - 1)) == 0 : false;  
3. } 
   

16. Get remainder of 2ⁿ

1. int quyu(int m,int n){
2.     return m & (n - 1);
   
3. } 
   

17. Get average value of two values

1. int getAverage(int x, int y){  
2.         return (x + y) >> 1;   
3. ｝ 
   

Another way:

1. int getAverage(int x, int y){  
2.         return ((x ^ y) >> 1) + (x & y);   
3. }  
   

18. Get the mth bit of a number n

1. int getBit(int n, int m){  
2.     return (n >> (m-1)) & 1;  
3. } 
   

19. Set the bit to 1 at mth bit of a number n

1. int setBitToOne(int n, int m){  
2.     return n | (1 << (m-1));  
3. } 
   

20. Set the bit to 0 at mth bit of a number n

1. int setBitToZero(int n, int m){  
2.     return n & ~(1 << (m-1));
   
3. } 
   

Here is one more article regarding using bit operation to realize addition, subtraction, multiplication and division. : Implementation of +,-,*,/ with bitwise operator

Source : http://blog.csdn.net/nash_/article/details/8262185