How to access the bit at a particular index? [closed]

Question

How do I access the bit at a particular index in a number using shift operator. Say the number is $n$.
Let the index be $i (i: 0 \le i < n).$
Rightmost bit is $0$.

The only thing I can think of is:
$def \, bitAt(n, i)$
$\,\,\,\,\,\,\,\,m = n >> i+1$
$\,\,\,\,\,\,\,\,v = n >> i$
$\,\,\,\,\,\,\,\,m = m << 1$
$\,\,\,\,\,\,\,\,if(m == v)$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,return \, 0$
$\,\,\,\,\,\,\,\,else$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,return \, 1$
$\,\,\,\,\,\,\,\,end$
$end$

Is there a more efficient way to do this?

Answer 1

This can be done by running an AND between n and a number with it's ith bit set to 1, which can be obtained by 1 << i. The result of this operation will be zero if the ith bit in n is zero, or 1 << i otherwise.

Case of zero:

n     = 10111
m     = 01000 (1 << 3)
n & m = 00000

Case of one:

n     = 10111
m     = 00100 (1 << 2)
n & m = 00100 (1 << 2)

So:

$m = 1 << i$

$\text{if}\; (n \;\&\; m == 0)$

$\;\;\;\;\text{return} \;0$

$\text{else}$

$\;\;\;\;\text{return} \;1$