Question

Can some one explain how it's get the first and second bit, so, how is (i&2)/2 getting the second bit ?
for(i=0; i < 4; i++) {
bit_a = (i & 2) / 2; // Get the second bit.
bit_b = (i & 1); // Get the first bit.
printf("%d | %d = %d\n", bit_a, bit_b, bit_a | bit_b);
}

Answer 1

bit_a=(i&2)/2;
This instruction can be written in C also(it's more appropriate to write it that way):
bit_a=(i&2)>>2;
Now at first iteration i=0:
i= 0000 0000
2= 0000 0010
-----&-------
i&2=0000 0000
2= 0000 0010
----->>------
bit_a=0000 0000//The second bit.You can't see it because it's equal to the first one=0
i= 0000 0000
1= 0000 0001
-----&--------
bit_b=0000 0000//The first bit.You can't see it because it's equal to the first one=0
now in printf you got
bit_a|bit_b=0000 0000
Now at second iteration i=1:
i= 0000 0001
2= 0000 0010
-----&-------
i&2=0000 0000
2= 0000 0010
----->>------
bit_a=0000 0000//The second bit pushed on the first position division with 2.
i= 0000 0001
1= 0000 0001
-----&--------
bit_b=0000 0001//The first bit.
now in printf you got
bit_a|bit_b=0000 0001
.
.
REVIEW BIT'S OPERATIONS CHAPTER I HOPE YOU GOT IT!

Answer 2

Let's just consider three bit numbers.
```
0 000
1 001
2 010
3 011
4 100
5 101
6 110
7 111
```
Some things should be coming clear now.
1 is the first bit `001`
2 is the second bit `010`
4 is the third bit `100`

Notice how 3 is `011` and 6 is `110`

Each time you multiply by 2, you shift everything to the left.
There is actually an operator in C that lets you shift everything to the left multiple times.

It's the `<<` operator. So `1 << 2` takes `1=001` shifts it `2` times so it becomes `100=4`. In general, `x << n` is a very fast way to calculate \(x \cdot 2^n\).

An even easier way to get the nth bit would be to use the left shift operator.
`(x >> (n -1)) & 1` will give you the nth bit. Let's disect it.

Suppose you have `7=111`. If you want the second bit, first it will do `n-1` which evaluates to 1. This means shift right once. We would shift 0 times for the first bit because it is already the right most bit. So it will do `111 >> 1` which will give us `011`. Next thing is the ` & 1`. This operation will and each bit together. So we have `011 & 001` which results in `001`.

Consider if you want the second bit of 5. Here is how it will work:
```
(5 >> (2 - 1)) & 1
(5 >> 1) & 1 // 5 = 101, 010 = 6
2 & 1 // 010 & 001 = 000
0
```
So the second bit of 5 doesn't exist, and we get 0.