Bit operation skills

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

(FormatOmittedModified)

1. Get the maximum value of int type

 
Int getmaxint () {return (1 <31)-1; // 2147483647. Brackets cannot be omitted due to priority}
 

Another method:

 
Int getmaxint () {return-(1 <-1)-1; // 2147483647}
 

Another method:

Int getmaxint () {RETURN ~ (1 <31); // 2147483647}
 

Another method:When the C language does not know how many bytes int occupies

 
Int getmaxint () {return (unsigned INT)-1)/2; // 2147483647}
 

2. Get the minimum value of int type

 
Int getminint () {return 1 <31; //-2147483648}
 

Another method:

 
Int getminint () {return 1 <-1; //-2147483648}
 

3 . Obtain the maximum value of the long type.

 
Long getmaxlong () {return (unsigned long)-1)/2; // 2147483647}
 

Obtain the minimum value of long, which is the same as the maximum value of other types. 4 . Multiply by 2

 
Int multwo (int n) {// calculate n * 2 Return n <1 ;}
 

5 . Divide by 2

 
Int divtwo (int n) {// return n> 1 cannot be used for negative odd values; // divided by 2}
 

6. Multiply by 2 to the power of m

 
Int multwopower (int n, int m) {// calculate n * (2 ^ m) return n <m ;}
 

7 . Divide by 2 to the power of m

 
Int divtwopower (int n, int m) {// calculate n/(2 ^ m) return n> m ;}
 

8 . Judge the parity of a number

 
Boolean isoddnumber (int n) {return (N & 1) = 1 ;}
 

9 . You do not need to exchange temporary variables for two numbers)

 
Void swap (int * a, int * B) {(* A) ^ = (* B) ^ = (* A) ^ = (* B );}
 

General version:

 
A ^ = B; B ^ = A; A ^ = B;
 

10. Take the absolute value (On some machines,Efficiency ratio n> 0
? N:-n High)

 
Int ABS (int n) {return (N ^ (n> 31)-(N> 31);/* n> 31 get the N symbol, if n is positive, N> 31 is equal to 0. If n is negative, N> 31 is equal to-1. If n is positive, N ^ 0 = 0, the number remains unchanged, if n is a negative number and N ^-1 needs to calculate the complement code of N and-1, and then perform an exclusive or operation, the result N is changed to the absolute value of N minus 1, minus-1 is the absolute value */}
 

11. Take the maximum value of two numbers (On some machines,Efficiency Ratio A> B? A: B high)

Int max (int A, int B) {return B & (a-B)> 31) | &(~ (A-B)> 31);/* If a> = B, (a-B)> 31 is 0, otherwise it is-1 */}
 

C LanguageVersion:

 
Int max (int x, int y) {return x ^ (x ^ y) &-(x <y);/* If x <Y x returns 1, otherwise, 0 is returned, and 0 is executed and the calculation result is 0, which is the same as-1 and the calculation result */}
 

12. Take the minimum value of two numbers (On some machines,Efficiency Ratio A> B
? B: A high)

General edition:

 
Int min (int A, int B) {return a & (a-B)> 31) | B &(~ (A-B)> 31);/* If a> = B, (a-B)> 31 is 0, otherwise it is-1 */}
 

C language:

Int min (int x, int y) {return y ^ (x ^ y) &-(x <y);/* If x <Y x returns 1, otherwise, 0 is returned, and 0 is executed and the calculation result is 0, which is the same as-1 and the calculation result */}
 

13. Determine if the symbols are the same

 
Boolean issamesign (int x, int y) {return (x ^ y)> 0; // true indicates that X and Y have the same symbol, and false indicates that X, Y has the opposite symbol. }
 

1 4. Calculate the N power of 2

Int getfactorialoftwo (int n) {// n> 0 return 2 <(n-1); // n of 2}
 

15. judge whether a number is a power of 2.

 
Boolean isfactorialoftwo (int n) {return (N & (n-1) = 0;/* if it is a power of 2, N must be 100... n-1 is 1111 .... therefore, the calculation result is 0 */}
 

16. Obtain the remainder from the nth power of 2

 
Int Quyu (int m, int N) {// n returns m to the power of 2 & (n-1);/* if it is a power of 2, N must be 100... n-1 is 1111 .... therefore, with the operation result, the non-0 bits of m in the N range are retained */}
 

17. Calculate the average of two integers

 
Int getaverage (int x, int y) {return (x + y)> 1 ;}
 

The following are the three most basic binary operations.18. From the low position to the high position, take the m position of N.

 
Int getbit (int n, int m) {return (n> m-1) & 1 ;}
 

19. From the low position to the high position. Place N at the M position 1.

 
Int setbittoone (int n, int m) {return n |... 1... 000 N is performing an OR operation with this number */}
 

20. From the low position to the high position, set the m position of N to 0.

Int setbittozero (int n, int m) {return N &~ (1 <m-1);/* shift 1 to the left and find the M-bit. After the reverse operation, the value is 111... 0... 1111 N and then perform operations with this number */}
 

AppendixProgramThere is no substantial improvement in the efficiency of bit operations, and some are also the knowledge of bit operations (interview may encounter) Calculation Calculate n + 1

 
-~ N
 

CalculationN-1

 
~ -N
 

Reverse number

 
~ N +1;
 

If (x = A) x = B; If (x = B) x =;

 
X = a ^ B ^ X;
 

SIGN function. The parameter is N. If n> 0, 1 is returned. If n <0,-1 is returned. If n = 0, 0 is returned.

 
Return !! N-(unsigned) n> 31) <1 );