Number of 1s in a binary string

[meher81]

I have a n-bit binary string. There are not more than two 1s in it. How can I found the simplest way to detect the number of 1's in it?
Thanks

 

[DineshSL]

Rotate the string one bit at a time; check n-th bit (or zeroth bit) if it is 1, increase the count. You have to rotate string n time to appear all bits in n-th position (on in zeroth position).

Thanks,

 

[meher81]

Thank you
But I want to find the solution without using any loop. Is there any logical statement for it?

 

[alexan_e]

You can also use a char pointer to check each char of the array which should be faster that rotate but without a loop I don't see a way to do what you want.

Alex

 

[westbend]

A general way to do this is the combinatorial loop below
Size m to handle the max value possible (generally log2+1)

integer count_i;
reg [m:0] count;
always @(*) begin
count = 0;
for (count_i=0; count_i<n; count_i = count_i + 'd1 )
count = count + vector[count_i]; // count 1s in vector
end

How efficient it is going to depend a lot on your synthesizer.

I would be interested in better ways to do this.