how is line 2 & 3 getting the second and first bit?

[futSecGuy1990]

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);  
 }

[weaknessforcats]

OK.

Let's assume i is 8 bits to keep things simple. Let's also assume i has a value of 245 which is 11110101 in binary.

The bits are numbered 0 to 7 from right to left:

11110101
..........^.... ... bit 2

Now you can AND this number with 00000100. This is called a mask.

11110101
00000100
--------------- AND
00000100

So the result of the AND will have a 1 when the number AND the mask both have a 1 in the same bit position.

In this case the result of the AND is equal to the mask. This means that bit 2 is 1 in the variable i.

unsigned int i = 245;
unsigned int mask = 4;
unsigned int result;

result = i & mask;   //The & is the bitwise AND operator

if (result == mask)
{
    //bit 2 is ON
}

[futSecGuy1990]

Ok, im still a little confused tho, sorry if this is frustrating For you!!!

 bit_a = (i & 2) / 2; // Get the second bit.

so keeping things simple. This is comparing
0000 0000
&
0000 0010 and the / 2 is saying look at the 2ND bit?

[futSecGuy1990]

Sorry I posted a reply but did not post to your answer. Could you check my reply out and see if I'm understanding this correctly

[weaknessforcats]

In this code:

bit_a = (i & 2) / 2; // Get the second bit.

it says get the second bit. 2 is 00000010. The 1 is in bit 1, not bit 2. It is bit 1 because the position of the bit is 2 to the 1st power.

The bit on the far right is bit 0. To the left of bit 0 is bit 1.

The bit is either ON (1) or OFF(0).


Now look at i & 2:


Let's look at the possible conditions

00000010

or

00000000

If you AND with 2 (00000010) you get:

00000010
00000010
............... ..AND
00000010

or you get

00000000
00000010
............... ..AND
00000000


In the first case the result of the AND is 00000010, which is 2. i & 2 becomes 2. So now you take 2 / 2 which is 1. Hence the bit is ON.

The other case is 00000000 which is 0. Then 0/2 is 0. Hence the bit is OFF.

[futSecGuy1990]

Ok so, let's say i =3

Bit_a
0000 0011
&
0000 0010
------------
0000 0010 = (2)/2 = 1
So on the final iteration
bit_a = 1

Bit_b
0000 0011
&
0000 0001
-----------
0000 0001 = 1
Bit_b = 1

And then in the print statement compares bit_a , bit_b using the OR operator

So...
0000 0001 bit_a
or
0000 0001 bit_b
------------
0000 0001 so the result of the printf in the final iteration should be 1 ?

Someone on another forum, as far as I understood it, was saying that the /2 is a different form of the shift operator, but it's really just the result of AND divided by 2. And the mask in this code would be the 2 for bit_a and 1 for bit_b. And the mask is used to look at a specific portion of 'i'. So in this case I'm looking at bit1 for the variable bit_a and bit0 for the variable bit_b.

[weaknessforcats]

Yes, /2 is not the shift operator. It is the division operator.

The shift operator is << or >> to shift bits left or right.

As to your analysis on bit_a or bit_b. it looks correct. Congratulations .

So far you have been testing bits. Using similar approaches, you can set or reset bits.