位操作（从X中提取第5位并在X中返回1位数）

Question

Assume X is an unsigned integer how will I be able to extract the 5th bit from X and return the number of 1 bits in X. Note: these two operations don't depend on each other.

For extracting the 5th bit from XI was thinking to use the & operation to do something like this: XXXX1X & 000010

I am not really sure how to do the return the number of 1 bits in X.

Thanks!

Answer 1

Counting the number of 1-bits is relatively straightforward; you iterate over all of the bits of your number and add to some accumulator the number of bits that are set:

unsigned int X;     
unsigned int count; 

for (count = 0; X; X >>= 1) {
   count += X & 1;
}

How it works:

1. Initialize count to 0
2. Start from the MSB (most significant bit) [this is the current bit]
3. Add to count the result of the current bit & 1
   • 1 if the bit is set
   • 0 if the bit is not set
4. Shift X to the right 1 bit, so the current bit is now the next MSB
5. Repeat step 3

Extracting the 5th bit is also straightforward, simply right shift your number X by 5 and compute the logical AND with 1:

unsigned int fifthBit (unsigned int X) {
   return (X >> 5) & 1;
}

Answer 2

A shorter and simpler way to find number of set bits in an int is following:

int num_1_bits(unsigned int x)  // or unsigned long x; or unsigned short x;
{
   int bitcount = 0;
   while(x)
   { 
      x &= (x - 1);
      ++bitcount;
   }
   return bitcount;
}

How does this work?

For an unsigned number x , the sequence of bits in x - 1 will be the complement of x till the first set bit of x counted from LSB.

ie for x = 20 [0b11100] ; x - 1 = 19 [0b11011]
Here Bit sequence of 19 is complement of 20 till first set bit in 20 (first set bit at position 3 from LSB).

Therefore,
if you do x & (x - 1) , this operation will consume one set bit from x .

Hence,
if you want to continue doing this until x becomes 0 , you will need to do it as many times as there are set bits in x , which in turn will give you the count of set bits in x .

This method is better because the loop will run only as many times as there are 1 bits in x and hence doesn't depend on datatype of x .

Answer 3

作为补充，我精确，如果你正在寻找的效率，也有一些内置的硬件指令来执行对位计算（数零：你可以推断出多少个1-bit ，你有那么）看到此文章作为例子。