Bitwise operators in c# OR(|), XOR(^), AND(&), NOT(~)

This post is a brush up on some of the bitwise operators that we shall be using for creating bitmasks to set permissions(my next post will be on exactly this subject. Bitmasks).

The c# bitwise operators I shall be covering here and are of particular interest to me in my followup post are :

1. binary OR(|) operator
2. binary AND(&) operator
3. XOR (^) operator
4. Not (~) operator

Binary OR(|) operator

Binary | operators are predefined for the integral types and bool. For integral types, | computes the bitwise OR of its operands. For bool operands, | computes the logical OR of its operands; that is, the result is false if and only if both its operands are false.

Lets focus on the integral types(types that can be represented as numbers) and how the computation takes place.

protected void Page_Load(object sender, EventArgs e)
{
     byte a = 7;
     byte b = 9;
     int orComputed = a | b;
     Response.Write(string.Format(
        "<br />{0} | {1} OR computed :{2}", 
               a, b, orComputed));

}

Output :
7 | 9 Result :15

To understand what just happened we need to look at bits and how each bit(which contains an integral type) is understood by the computer. The computer is going to understand this as binary data in the form of 0's and 1's eg. -->> 00101010(which represents the number 42), this is actaully much easier for a computer to handle since each bit(or integral type as in number), can be interpreted as either on or off signal.

0 = off
1 = on

To understand the bitwise operator with clarity and how it computes the result, lets first construct a base 2 table from right to left incrementing the exponent by 1 for each power. So, since we have 8 bits in a byte, we make a table that lists 8 elements 128, 64, 32, 16, 8, 4, 2, 1.

128	64	32	16	8	4	2	1	Result
0	0	0	0	0	0	0	1	1
0	0	0	0	0	0	1	0	2
0	0	0	0	0	1	0	0	4
0	0	0	0	1	0	0	0	8
0	0	0	1	0	0	0	0	16
0	0	1	0	0	0	0	0	32
0	1	0	0	0	0	0	0	64
1	0	0	0	0	0	0	0	128

If you will look at the above pattern more closely, we notice how the 1 keeps moving 1 digit to the left for every new value in binary. This is also known as a "bit shift".

Ok, now that we have our binary 8bit table with digits to the power of 2^ up to 128, so for a simple 7 | 9, which yielded the result 15, lets look at how it was computed by first creating the binary representation using our base 2 table and then computing the binary | operation, so :

128	64	32	16	8	4	2	1	Calculation	Result
0	0	0	0	0	1	1	1	(4+2+1)	7
0	0	0	0	1	0	0	1	(8+1)	9
0	0	0	0	1	1	1	1	(8+4+2+1)	15

As you can see from the above, binary (|) is being performed on each pair of corresponding bits. If a 1 is present in the either bit or in both bits, otherwise it's 0. Not clear ? Lets try that again. How did we get the result 1111 ? It's a simple comparision. If either bit stacked ontop of one another are 1 or both are 1, then the result is 1, otherwise the result is 0. So 1 on 1 = 1, 1 on 0 = 1 but 0 on 0 = 0.

Binary AND(&) operator

Quoting MSDN Docs :

Binary & operators are predefined for the integral types and bool. For integral types, & computes the logical bitwise AND of its operands. For bool operands, & computes the logical AND of its operands; that is, the result is true if and only if both its operands are true.

So, lets quickly test this :

protected void Page_Load(object sender, EventArgs e)
{
     byte a = 7;
     byte b = 9;
     int orComputed = a & b;
     Response.Write(string.Format(
            "<br />{0} & {1} Result :{2}", 
                    a, b, orComputed));

}

Output is :
7 & 9 Result :1

Lets convert to binary using our base 2 table, and dig deeper  :

128	64	32	16	8	4	2	1	Calculation	Result
0	0	0	0	0	1	1	1	(4+2+1)	7
0	0	0	0	1	0	0	1	(8+1)	9
0	0	0	0	0	0	0	1	(1)	1

so, how did we get the binary result 0001(1) ?  binary (&) is being performed on each pair of corresponding bits. The comparision is made based on both bits stacked ontop of one another. if both are one's, then we have a 1, anything else is a 0. so 1 on 1 = 1, 1 on 0 = 0, 0 on 0 = 0.
Binary Xor (^) operator

Quoting MSDN Docs:

Binary ^ operators are predefined for the integral types and bool. For integral types, ^ computes the bitwise exclusive-OR of its operands. For bool operands, ^ computes the logical exclusive-or of its operands; that is, the result is true if and only if exactly one of its operands is true.

A simple c# example :

protected void Page_Load(object sender, EventArgs e)
{
     sbyte a = 7;
     sbyte b = 9;
     int orComputed = a^b;//on a negative
     Response.Write(string.Format(
              "<br />{0} ^ {1} Result :{2}", 
                 a,b, orComputed.ToString()));

}

output :
7 ^ 9 Result :14

Now, how did binary (^) Xor'ing 7 ^ 9 produce 14 ? This is because binary (^) is being performed on each pair of corresponding bits. If we have two matching zero's or one's, then the result is 0, otherwise 1. SO, 1 on 1 = 0, 0 on 0 = 0, 1 on 0 = 1, 0 on 1 = 1. Lets look at an example in our base 2 table of 7 and 9 in binary format :

128	64	32	16	8	4	2	1	Calculation	Result
0	0	0	0	0	1	1	1	(4+2+1)	7
0	0	0	0	1	0	0	1	(8+1)	9
0	0	0	0	1	1	1	0	(8+4+2)	14

Not (~) operator

Quoting MSDN Docs:

The ~ operator performs a bitwise complement operation on its operand, which has the effect of reversing each bit. Bitwise complement operators are predefined for int, uint, long, and ulong.

This is also a unary operator so we don't need to pass a value. Unary operators perform an operation on a single operand eg. ~n and not x~y.

Binary (~) operator is the most confusing one. It performs reversing of each bit, so reversing a positive can produce a negative value. Negative Values Use Two's Complement format and is quite tricky.

A simple example :

protected void Page_Load(object sender, EventArgs e)
{
     sbyte a = 7;
     int orComputed1 = ~-a;//on a negative
     int orComputed2 = ~a;// on a positive
     Response.Write(string.Format(
                "<br />~-{0} Result :{1}", 
                    a, orComputed1));


     Response.Write(string.Format(
             "<br />~+{0} Result :{1}", 
                    a, orComputed2));


}

output :
~-7 Result :6
~+7 Result :-8

As you can see, when the NOT(~) operator is applied to a positive number, the resulting value is a negative. This is because the value is reversed. Since this reversing yields a negative or positive depending on what value your reversing, use a signed type (a type that can store a negative value).

In the base table below, our binary number gets inverted. all 1's become 0 and 0 becomes 1.

A signed binary uses the left most bit to keep track of the sign(negative or positive).
In the c# compiler, negative values are represented internally in two's complement format. Two's complement can be obtained by negating each bit of the value, then adding 1. Performing two's complement twice generates the original value.

In the following base 2 table, one typical gotcha when inverting is to remember to start the inversion from the left of the first 1 value in our binary. So, say had we a binary representation of the digit 7(00000111), then we start inverting at the 2nd digit starting at the right, skipping the first(since it's 1)  and get 11111001. A last thing to also note is how in our calculation we added a negative -1 to our result instead of a positive 1 -->> (-128+64+32+16+8+1)-1 ; While the two's complement format states that we need add 1, we are actually deducting 1 versus adding 1. This is because we have to consider 0 in the equation.

An easy method to get the two's complement of +7 :(left most is used to track sign, so add -1 to the result) :

128	64	32	16	8	4	2	1	Calculation	Result
0	0	0	0	0	1	1	1	(4+2+1)	7
1	1	1	1	1	0	0	1	(-128+64+32+16+8+1)-1	-8

Note the left most significant pair of corresponding bits in red. A value of 0 means positive and a value of 1 means negative and is used for tracking the sign in a signed binary.

One question still remains, why is the two's complement always a positive less(6) and a negative more(-8) ?
Why aren't we getting -7 for a 7 and vice versa. I guess the answer to this is simply "logic", we have to consider the 0.

-8	-7	-6	-5	-4	-3	-2	-1
7	6	5	4	3	2	1	0