How to find a binary number all 1s positions use bit manipulation?

Question

For example, I have a number 210 or 11010010 in binary, the 2nd 5th 7th and 8th position is 1, so the result is [2, 5, 7, 8].

What is the fastest way to find that?

Please pick *one* programming language, the one you're working with. — CertainPerformance, Oct 28 '18 at 10:45
Possible duplicate of [What are Python's equivalent of Javascript's reduce(), map(), and filter()?](https://stackoverflow.com/questions/31127824/what-are-pythons-equivalent-of-javascripts-reduce-map-and-filter) — mplungjan, Oct 28 '18 at 10:46

score 3 · Accepted Answer · answered Oct 28 '18 at 10:51

3

Convert the number to binary:
```
>>> format(210, 'b')
'11010010'
```

Use a list comprehension and enumerate to find the indices of the 1s:

>>> [i for i, digit in enumerate(reversed('11010010'), 1) if digit == '1'] 
[2, 5, 7, 8]

answered Oct 28 '18 at 10:51

Aran-Fey

39,665
11
104
149

score 0 · Answer 2 · answered Oct 28 '18 at 10:52

0

Just loop over your binary number to get the result.

a=format(210, 'b')
x=[]
for i in range(0,len(a)):
    if a[i] == '1':
        x.append(i+1)

1 position is [1, 2, 4, 7]

answered Oct 28 '18 at 10:52

Ankur Gulati

291
1
12

score 0 · Answer 3 · answered Oct 28 '18 at 11:12

The fastest way is to precompute all combinations (only 256 of them) by whatever method, and store in a list (of lists).

You can build the list by appending 1 to all entries, then 2 to every other entries, then 3 to every fourth entries, and so on.

score 0 · Answer 4 · answered Oct 28 '18 at 14:28

A solution in C++.

#include<iostream>
using namespace std;
int main()
{
    int n=10;
    int counter=1;
    while(n)
    {
        if(n&1==1)
        {
            cout<<counter<<" ";
        }
         n=n>>1;
        counter++;
    }

}

Read a bit about bitwise shift and & operator, to understand the solution properly.

How to find a binary number all 1s positions use bit manipulation?

4 Answers4