How does a bit shift work?

Explanation about bit shift.

Used languages: JavaScript, C#.

Shift to right.

Number: 93.

Step 0.

We have a number 93, in binary system it is 1011101.

JS:

console.log( parseInt(1011101, 2) ) // -> 93.

C#:

Console.WriteLine( Convert.ToInt32("1011101", 2) ) // -> 93.

Step 1.

Due to using operator ">>" in expression 93 >> 1, we tell compiler to do a bit shift for 1 symbol, thus or number 93 will be transformed into 46.

JS:

console.log( parseInt(101110, 2) == 93 >> 1) // -> true.

C#:

Console.WriteLine( 93 >> 1 == Convert.ToInt32("101110", 2) ) // -> true.

**Clarification **:

1011101 -> 93. 101110 -> 46.

All existing numbers have to be shifted into right side for 1 sign. So, it means that 1st number will be on the second place, 2nd on the 3rd etc, but the last number leaves collection. As you see we had a last number "1", but it have to be removed.

Step 2.

93 >> 2 == 46 >> 1 -> 23.

Step 3.

93 >> 3 == 46 >> 2 == 23 >> 1 -> 11.

Step 4.

93 >> 4 == 46 >> 3 == 23 >> 2 == 11 >> 1 -> 5.

Step 5.

93 >> 5 == 46 >> 4 == 23 >> 3 == 11 >> 2 == 5 >> 1 -> 2.

Step 6.

93 >> 6 == 46 >> 5 == 23 >> 4 == 11 >> 3 == 5 >> 2 == 2 >> 1 -> 1.

Step 7.

93 >> 7 == 46 >> 6 == 23 >> 5 == 11 >> 4 == 5 >> 3 == 2 >> 2 == 1 >> 1 -> 0.

So, we had a 7 iterations, thereafter the last received sign is zero. All next bit shift operation can't change that and we'll have a sign 0 all the time.

Shift to left.

It is completely opposite to previous method. Compiler will add a 1 sign into the start, which increases the number.

Number: 1.

Step 0.

How we did that? First of all, you have to understand the principle of representation the numbers in binary system. See a table below:

Metric system	Binary system
0	0
1	1
2	10
3	11
4	100
5	101
6	110

etc .

So, when we use operator << and apply it for number 1, the result will be 2 because we have a bits shift to left.

Step 1.

1 << 1 -> 2.

Step 2.

2 << 1 -> 4.

Step 3 (a consistent shift).

1 << 1 << 1 -> 4.

Let's check that in C# compiler:

Console.WriteLine(Convert.ToString(4, 2)); // get a binary system representation
Console.WriteLine(Convert.ToInt32("100", 2)); // convert number into metric system
Console.WriteLine(1 << 1 << 1 == Convert.ToInt32("100", 2)); // compare results

Betwise AND (&).

Now let's talk about common expression task - betwise and. According to code above, there is no a new principles, just only another representational behaviour.

Supposably, we want to use & operator to numbers 5 and 7. So, here is a bits transformation:

Metric system	Position 3	Position 2	Position 1
5	1	0	1
7	1	1	1
5	1	0	1

Let's change numbers and now we're having 9 and 5:

Metric system	Position 4	Position 3	Position 1
9	1	0	1
5	0	1	1
1	0	0	1

Check it using JavaScript:

console.log( parseInt(101, 2) & parseInt(111, 2) ); // -> 5
console.log( parseInt(1001, 2) & parseInt(101, 2) ); // -> 1

Betwise OR (|).

In this case all combinations return 1 except 0 & 0. Example:

Position 1	Position 2	Result
0	0	0
0	1	1
1	0	1
1	1	1

Supposably, we use OR operator with numbers 7 and 5, so compiler does the following actions:

Metric system	Position 3	Position 2	Position 1
5	1	0	1
7	1	1	1
7	1	1	1

Betwise Exclusive OR (^).

Only different signs return "1", so there are [0 - 1], [1 - 0] combinations, in other cases will returned "0".

Metric system	Position 3	Position 2	Position 1
5	1	0	1
7	1	1	1
2	0	1	0