Counting number of bit sets in an integer (Hamming Weight)

There is always a multiple ways to get a solutions for given problem and I am not going to argue this is the best.

int count_number_of_bit_set(unsigned value) {

    unsigned int shift[] = {1, 2, 4, 8, 16};

    unsigned int magic_number[] = {

                                   0x55555555,

                                   0x33333333,

                                   0x0F0F0F0F,

                                   0x00FF00FF,

                                   0x0000FFFF};

    unsigned char i = 0;

    for (; i <= 4; i++) {

         value = (value & magic_number[i]) +

                 ((value >> shift[i]) & magic_number[i]);

    }

    return value;

}

Let understand the logic inside the code,

The above picture is drawn to explain the concept for a byte. which can be extended for multiple bytes by applying the above concept.

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