Bit operation summary

1. Only integer operations can be performed, such as char, short, int, and long (whether signed or unsigned)

 

2. >>,< the priority is smaller than the arithmetic operator, so the parentheses in the expression a = B + (c> 1) cannot be less. Note that the right shift> operator,For the unsigned number, the left side is filled with 0, while for the signed number, it may be filled with 0 or "symbol bit". Therefore, when we always want to fill the number with 0, it must be declared with unsigned (left shift <always fills with 0, whether it is a signed number or an unsigned number)

 

3. Several examples of The C Programming language are classic:

 /* Obtain the n bits starting from position P */
Unsigned int getbits (unsigned x, int p, int n)
{
Return (x> (p + 1-n ))&(~ (~ 0 <n ));
}

/* Reverse the n bits starting from position p, And the rest remain unchanged */
Unsigned int invert (unsigned x, int p, int n)
{
Return (x ^ (~ (~ 0 <n) <(p + 1-n )));
}
11011001
^ 00111100 (just read this)

 

4. In the beauty of programming, the bitwise operation of Question 1 in Chinese chess is also very concise and beautiful, and it is recorded here

 # Define HALF_BITS_LENGTH 4
# Define FULLMASK 255

# Define LMASK (FULLMASK <HALF_BITS_LENGTH) // 11110000
# Define RMASK (FULLMASK> HALF_BITS_LENGTH) // 00001111

# Define RSET (B, n) (B = (B & LMASK) | n) // reset n to the right
# Define LSET (B, n) (B = (B & RMASK) | (n <HALF_BITS_LENGTH) // reset n on the left

# Define RGET (B) (B & RMASK) // obtain the right value
# Define LGET (B) (B & LMASK)> HALF_BITS_LENGTH) // obtain the left Value

It is easy to modify the above slightly to fit any bit.

 

5. The following content is reproduced from: http://blog.sina.com.cn/s/blog_66ad7bba0100hf9k.html

 Q: What should I do if I want to remove the first occurrence 1 in a binary number, such as 6-> 4, 14-> 12, 5-> 4?
Answer: The fastest algorithm: n & (n-1)

The problem is, what should I do if I want to obtain the first occurrence 1 in a binary digit, such as 6-> 2, 7-> 1, 12-> 4?
Answer: Likewise, you can get n-(n & (n-1 ))

Problem: judge whether a number is a power of 2.
Answer: (n & (n-1) = 0 & n> 0

Problem: Write the sign function. The parameter is int n. If n is negative, return-1. If positive, return 1. 0. Return 0.
Solution 1: return !! N-(unsigned) n> 31) <1)
Solution 2: return !! N + (n> 31) * 2 (this is not necessarily true for all compilers)

Problem: to calculate the average of two integers, consider overflow.
A: This calculation method ensures that it is not affected by the overflow problem: (x & y) + (x ^ y)> 1)

Problem: Calculate a power greater than n and the smallest power of 2, such as 3-> 4, 6-> 8,100-> 128,256-> 512
Answer:
Int calc (int n)
{
N | = n> 1;
N | = n> 2;
N | = n> 4;
N | = n> 8;
N | = n> 16;
Return n + 1;
}