Counting the number of bits that are set

Question

I want to count the number of bits in a binary number that are set. For example, user enter the number 97 which is 01100001 in binary. The program should give me that 3 bits are

Answer 1

Since this sounds like homework, I'm not going to give the MIPS code, but here's the algorithm (written in C). It should be straightforward to translate into MIPS:

int bits(unsigned int number)
{
    // clear the accumulator
    int accumulator = 0;
    // loop until our number is reduced to 0
    while (number != 0)
    {
        // add the LSB (rightmost bit) to our accumulator
        accumulator += number & 1;
        // shift the number one bit to the right so that a new
        // bit is in the LSB slot
        number >>= 1;
    }
    return accumulator;
}