Bit manipulation operators manipulate individual bits within a variable.
Why bother with bitwise operators?
In the past, memory was extremely expensive, and computers did not have much of it. Consequently, there were incentives to make use of every bit of memory available. Consider the bool data type — even though it only has two possible values (true and false), which can be represented by a single bit, it takes up an entire byte of memory! This is because variables need unique addresses, and memory can only be addressed in bytes. The bool uses 1 bit and the other 7 go to waste.
Using bitwise operators, it is possible to write functions that allow us to compact 8 booleans into a single byte-sized variable, enabling significant memory savings at the expense of more complex code. In the past, this was a good trade-off. Today, it is not.
Now memory is significantly cheaper, and programmers have found that it is often a better idea to code what is easiest to understand and maintain than what is most efficient. Consequently, bitwise operators have somewhat fallen out of favor, except in certain circumstances where maximum optimization is needed (eg. scientific programs that use enormous data sets, or games where bit manipulation tricks can be used for extra speed). Nevertheless, it is good to at least know about their existence.
There are 6 bit manipulation operators:
| Operator | Symbol | Form | Operation |
|---|---|---|---|
| left shift | << | x << y | all bits in x shifted left y bits |
| right shift | >> | x >> y | all bits in x shifted right y bits |
| bitwise NOT | ~ | ~x | all bits in x flipped |
| bitwise AND | & | x & y | each bit in x AND each bit in y |
| bitwise OR | | | x | y | each bit in x OR each bit in y |
| bitwise XOR | ^ | x ^ y | each bit in x XOR each bit in y |
Note: In the following examples, we will largely be working with 4-bit binary values. This is for the sake of convenience and keeping the examples simple. In C++, the number of bits used will be based on the size of the data type (8 bits per byte).
Left shift and right shift operator
The bitwise left shift (<<) shifts operator bits to the left. For example, consider the number 3, which is binary 0011. Left shifting by 1 (3 << 1) changes 0011 to 0110, which is decimal 6. Note how each bit moved 1 place to the left. Left shifting by 2 (3 << 2) changes 0011 to 1100, which is decimal 12. Left shifting by 3 (3 << 3) changes 0011 to 1000. Note that we shifted a bit off the end of the number! Bits that are shifted off the end of the binary number are lost.
The bitwise right shift (>>) operator shifts bits to the right. Right shifting by 1 (3 >> 1) changes 0011 to 0001, or decimal 1. The rightmost bit shifted off the end and was lost!
Although our examples above are shifting literals, you can shift variables as well:
unsigned int nValue = 4; nValue = nValue << 1; // nValue will be 8
Rule: When dealing with bit operators, use unsigned variables.
Programs today typically do not make much use of the bitwise left and right shift operator in this capacity. Rather, you tend to see the bitwise left shift operator used with cout in a way that doesn’t involve shifting bits at all! If << is a bitwise left shift operator, then how does cout << "Hello, world!"; print to the screen? The answer is that cout has overridden (replaced) the default meaning of the << operator and given it a new meaning. We will talk more about operator overloading in a future section!
Bitwise NOT
The bitwise NOT operator (~) is perhaps the easiest to understand of all the bitwise operators. It simply flips each bit from a 0 to a 1, or vice versa. Note that the result of a bitwise NOT is dependent on what size your data type is! For example, with 4 bits, ~4 (0100 binary) evaluates to 1011, which is 11 in decimal. In an 8 bit data type (such as an unsigned char), ~4 (represented as ~0000 0100) evaluates to 1111 1011, which is 251 in decimal!
Bitwise AND, OR, and XOR
Bitwise AND (&) and bitwise OR (|) work similarly to their logical AND and logical OR counterparts. However, rather than evaluating a single boolean value, they are applied to each bit! For example, consider the expression 5 | 6. In binary, this is represented as 0101 | 0110. To do (any) bitwise operations, it is easiest to line the two operands up like this:
0 1 0 1 // 5 0 1 1 0 // 6
and then apply the operation to each column of bits. If you remember, logical OR evaluates to true (1) if either the left or the right or both operands are true (1). Bitwise OR evaluates to 1 if either bit (or both) is 1. Consequently, 5 | 6 evaluates like this:
0 1 0 1 // 5 0 1 1 0 // 6 ------- 0 1 1 1 // 7
Our result is 0111 binary (7 decimal).
We can do the same thing to compound OR expressions, such as 1 | 4 | 6. If any of the bits in a column are 1, the result of that column is 1.
0 0 0 1 // 1 0 1 0 0 // 4 0 1 1 0 // 6 -------- 0 1 1 1 // 7
1 | 4 | 6 evaluates to 7.
Bitwise AND works similarly. Logical AND evaluates to true if both the left and right operand evaluate to true. Bitwise AND evaluates to true if both bits in the column are 1) Consider the expression 5 & 6. Lining each of the bits up and applying an AND operation to each column of bits:
0 1 0 1 // 5 0 1 1 0 // 6 -------- 0 1 0 0 // 4
Similarly, we can do the same thing to compound AND expressions, such as 1 & 3 & 7. If all of the bits in a column are 1, the result of that column is 1.
0 0 0 1 // 1 0 0 1 1 // 3 0 1 1 1 // 7 -------- 0 0 0 1 // 1
The last operator is the bitwise XOR (^), also known as exclusive or. When evaluating two operands, XOR evaluates to true (1) if one and only one of it’s operands is true (1). If neither or both are true, it evaluates to 0. Consider the expression 6 ^ 3:
0 1 1 0 // 6 0 0 1 1 // 3 ------- 0 1 0 1 // 5
It is also possible to evaluate compound XOR expression column style, such as 1 ^ 3 ^ 7. If there are an even number of 1 bits in a column, the result is 0. If there are an odd number of 1 bits in a column, the result is 1.
0 0 0 1 // 1 0 0 1 1 // 3 0 1 1 1 // 7 -------- 0 1 0 1 // 5
Bitwise assignment operators
As with the arithmetic assignment operators, C++ provides bitwise assignment operators in order to facilitate easy modification of variables.
| Operator | Symbol | Form | Operation |
|---|---|---|---|
| Left shift assignment | <<= | x <<= y | Shift x left by y bits |
| Right shift assignment | >>= | x >>= y | Shift x right by y bits |
| Bitwise OR assignment | |= | x |= y | Assign x | y to x |
| Bitwise AND assignment | &= | x &= y | Assign x & y to x |
| Bitwise XOR assignment | ^= | x ^= y | Assign x ^ y to x |
For example, instead of writing nValue = nValue << 1;, you can write nValue <<= 1;.
Summary
Summarizing how to evaluate bitwise operations utilizing the column method:
When evaluating bitwise OR, if any bit in a column is 1, the result for that column is 1.
When evaluating bitwise AND, if all bits in a column are 1, the result for that column is 1.
When evaluating bitwise XOR, if there are an odd number of 1 bits in a column, the result for that column is 1.
Quiz
1) What does 0110 >> 2 evaluate to in binary?
2) What does 5 | 6 evaluate to in decimal?
3) What does 5 & 6 evaluate to in decimal?
4) What does 5 ^ 6 evaluate to in decimal?
Quiz answers
 4.1 -- Blocks (compound statements) and local variables
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 Index
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 3.7 -- Converting between binary and decimal
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you put in the table that AND, OR, and XOR flip all digits in x but they do not.
[ Aah, the perils of copy and paste. :) Fixed. -Alex ]
When you do the compound XOR expression the result 0101 would equal 5 not 1.
[ Fixed. -Alex ]
What’s the difference in these: 3 << 1 and 3 <<= 1?
From what I understand, the end answer is the same, but in the second case the 3 itself changes to 6 (3 becomes bitwise shifted to the left 1, which is decimal 6). Whereas in the first case 3 stays 3 even after the operation; you’d need an x = 3 << 1 for example for it to have any context.
3 << 1 returns 6 whereas 3 <<= 1 does not compile since “3″ is not a variable name…
int x = 3; x = x << 1 ; and int x = 3 ; x <<= 1 ; are equivalent; eventually x = 6.
x <<= y is the same as x = x << y. Because of this, 3 <<= 1 is the same as 3 = 3 << 1, which obviously makes no sense since you can’t assign a value to the number 3!
Phun to look at ASCII table and how it is built up regarding bitwise operations.
I.e. Converting from upper to lower case.
A=65=01000001
a=97=01100001
32=00100000
So 65^32 (or A^32) gives 97 (or a).
So 97^32 (or a^32) gives 65 (or A).
#include <iostream>
int main()
{
using namespace std;
char chX,chX32;
for(int i=65;i<91;i++)
{
chX = i;
chX32 = i^32;
cout << i << ":t" << chX << "tt" << int(chX32) << ":t" << chX32 << "tt";
chX32 = chX32^32;
cout << int(chX32) << ":t" << chX32 << endl;
}
return 0;
}
I couldn’t understand the above comment.
Why he has mentioned of the digit 32 ?
Please make it clear…
In ASCII, upper case and lower case letters are separated by 32 bits. As he says, 65=’A', 97=’a’. ‘a’-'A’ = 32. By XORing 32 onto a lower or upper case letter, we can convert it from lower case to upper case or vice versa.
please,..is it possible to give me a usage (or compare use of bitwise operands and normal byte variables)? I understand the bit operation, but can not fing the ‘added value’ (for example in gaming).
thanks…
thanks to you alex, i learned the operation of bitwise operators but i dont know where can i use it. can you give me a site of example for program application.
I’m looking to create a new PRNG, and I need to be able to assign individual bits within a byte… Is there a way to decide what each bit will be within a byte without changing the other bits, and without finding out what symbol (letter/number/other) uses the specific binary code I need and then assigning it?
Hey Alex, This site is great. Do you know any site that with give an example to try this stuff out.
Thanks alot.
An easy example is in things like compression.. you can easily add 0’s and 1’s to things like characters.
An easy use to display the <> operators:
You have a character. Display its binary. Using bitwise <>, it is quite easy to.
char decode = (somesortofcharactefgettingthinghere);
char tempChar;
for (int i = 0; i < 8; i++) // we are using a character. It is 1 byte, or 8 bits (or is on my computer).
{
tempChar = decode
tempChar << i;
tempChar >> 7; // because bits are lost, we can seperate 0’s and 1’s by themselves.
cout << char;
}
I can’t easily explain that, but if you think it out in your head…
lets say our character is the number 97 (I think thats A, but it does not matter. It is only outputted as A, internally it is a number)
0 1 1 0 0 0 0 1
Alright, in our loop, we start by shifting it 0 units to the left. we then shift it to the right 7.
Our result? 0. Next time, move it 1 unit to the left, 7 to the right. 1. So on and so forth. You could have alternatively used a calculation, but this was just an example.
How about a simple encryption program? Have a secret list of passwords in a text file? Render it useless to hackers with a simple changing of 0’s to 1’s, and 1’s to 0’s! (this both encrypts and unencrypts it):
#include fstream
#include iostream
using namespace std;
int main()
{
char choice[250];
fstream file;
cout <> choice; //EDIT: this is wtf scrweed up by the site.. youi know how to use cin I assume >.>
file.open(choice, ios::binary | ios::in);
if (!file.is_open())
return 0;
char *data;
long begin, end, size;
file.seekg(0, ios::beg);
begin = file.tellg();
file.seekg(0, ios::end);
end = file.tellg();
size = end – begin;
data = new char[size];
file.seekg(0, ios::beg);
file.read(data, size);
file.close();
for (long i = 0; i < size; i++)
data[i] = ~data[i];
file.open(”output.woo”, ios::out | ios::binary);
file.seekp(0, ios::beg);
file.write(data, size);
return 0;
}
So if you did (7 >> 2), you get from 0111 to 0001, which is the number 1. Now how can you get back from that to 7? So the main use for this is so noone understands it except for the person who changed the value? To me, these bitwise operators seem useless except for Bitwise Not (~). I really do not understand what good you can get out of doing 5|6.
i had read a c++ book..i dint have this kind of satisfaction after reading the book.. its truly amazing..thank you guru alex.. god bless you..