Write a C function to return the number of set bits in an integer.

Question

Brad Murry · Accepted Answer

32 bits in an integer(int) data type on most systems.

Anonimo · Answer

This looks mos efficient,
int GetNumSetBits( int anInt )
{
 int count ;
while(anInt)
{
   count += (anInt&0x1);
   anInt >>= 1;
}
return count;
}

More efficient · Answer

while(n)
{
 n = n&(n-1)
count++;
}
return count;

cwneom78@gmail.com · Answer

#define BITS_IN_BYTE

int numSetBits (int

cwneom78@gmail.com · Answer

// C language implementation, handles different integer sizes
#define BITS_IN_BYTE    8L
#define LS_BIT_MASK     1L

int numSetBits (int intNumber) {
   int numBits = 0;
   for (int i = 0; i >= 1;                // right shift LS bit out
   }
}

Dave · Answer

int GetNumSetBits( int anInt )
{
   int count = 0;
   for( sizeof( int) )
   {
      count += (anInt & 0x01);
      anInt >>= 1;
   }
   return count;
}

shy · Answer

static const int NumberOfSetBits[8] = { /* in nibble, half byte */
  0, /* 0x0, b0000 */
  1, /* 0x1, b0001 */
  1, /* 0x2, b0010 */
  2, /* 0x3, b0011 */
  1, /* 0x4, b0100 */
  2, /* 0x5, b0101 */
  2, /* 0x6, b0110 */
  3  /* 0x7, b0111 */
  1  /* 0x8, b1000 */
  2  /* 0x9, b1001 */
  2  /* 0xA, b1010 */
  3  /* 0xB, b1011 */
  2  /* 0xC, b1100 */
  3  /* 0xD, b1101 */
  3  /* 0xE, b1110 */
  4  /* 0xF, b1111 */
};

#define NumBytesInInt 4 /* assumes 32bit system */

int GetNumberOfSetBits(int anInteger)
{

    uint8_t buf[NumBytesInInt ]; 
    int i;
    int count = 0;

    memcpy(buf, &anInteger, sizeof(buf));
    for (i=0; i> 4];
    }

    return count;
}

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