Fastest way to count number of 1s in a register, ARM assembly

So I had an interview question before regarding bit manipulation. The company is a well known GPU company. I had very little background in assembly language (weird despite being a phd student in computer architecture) and as this narrative would indicate, I botched it. The question was a simple:

"Write a fast code that will count the number of 1's in a 32-bit register."

Now I am in the process of studying arm assembly. So naturally I revisited this problem again and have come up with this code just by studying the ISA.

For you arm experts out there, is this correct? Is there a faster way to do this? Being a beginner, I naturally think this is incomplete. The AND instruction in "xx" feels redundant but there is no other way to shift a register in ARM isa...

R1 will contain the number of bits at the end while R2 is the register with bits we want to count. r6 is just a dummy register. Comments are enclosed in ()

    MOV   R1, #0                (initialize R1 and R6 to zero)
    MOV   R6, #0        
xx: AND   R6, R6, R2, LSR #1    (Right shift by 1, right most bit is in carry flag)
    ADDCS R1, #1                (Add #1 to R1 if carry  flag is set)
    CMP R2, #0                  (update the status flags if R2 == 0 or not)
    BEQ xx                      (branch back to xx until R2==0)

Solution

You could use a precomputed look up table and reduce the number of iterations to 2 or 4.

You could also use the logarithmic approach.

For more info see this Wikipedia article.