The bitwise operators

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// syntax test int i{ 1234'5678 }; char ch{ 'x' }; std::string str{ "Hello \'world\'" }; char ch2{ '\'' }; |

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// syntax test 2 char ch{ '\'' }; // awd |

C++ provides 6 bit manipulation operators, often called bitwise 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 |

Author's 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 actual programs, the number of bits used is based on the size of the object (e.g. a 2 byte object would store 16 bits).

For readability, we’ll also omit the 0b suffix outside of code examples (e.g. instead of 0b0101, we’ll just use 0101).

Bitwise left shift (<<) and bitwise right shift (>>) operators

The bitwise left shift (<<) operator shifts bits to the left. The left operand is the expression to shift the bits of, and the right operator is an integer number of bits to shift left by.

So when we say `x << 1`

, we are saying "shift the bits in the variable x left by 1 place". New bits shifted in from the right side receive the value 0.

0011 << 1 is 0110

0011 << 2 is 1100

0011 << 3 is 1000

Note that in the third case, we shifted a bit off the end of the number! Bits that are shifted off the end of the binary number are lost forever.

The bitwise right shift (>>) operator shifts bits to the right.

1100 >> 1 is 0110

1100 >> 2 is 0011

1100 >> 3 is 0001

Note that in the third case we shifted a bit off the right end of the number, so it is lost.

Here's an example of doing some bit shifting:

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#include <iostream> #include <bitset> int main() { std::bitset<4> x { 0b1100 }; std::cout << x << '\n'; std::cout << (x >> 1) << '\n'; // shift right by 1, yielding 0110 std::cout << (x << 1) << '\n'; // shift left by 1, yielding 1000 return 0; } |

This prints:

1100 0110 1000

Note that the results of applying the bitwise shift operators to a signed integer are compiler dependent prior to C++20.

Warning

Prior to C++20, don't shift a signed integer (and even then, it's probably still better to use unsigned)

What!? Aren't operator<< and operator>> used for input and output?

They sure are.

Programs today typically do not make much use of the bitwise left and right shift operators to shift bits. Rather, you tend to see the bitwise left shift operator used with std::cout to output text. Consider the following program:

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#include <iostream> int main() { unsigned int x { 0b0100 }; x = x << 1; // use operator<< for left shift std::cout << std::bitset<4>(x); // use operator<< for output return 0; } |

This program prints:

1000

In the above program, how does operator<< know to shift bits in one case and output *x* in another case? The answer is that std::cout has **overloaded** (provided an alternate definition for) operator<< that does console output rather than bit shifting.

When the compiler sees that the left operand of operator<< is std::cout, it knows that it should call the version of operator<< that std::cout overloaded to do output. If the left operand some other type, then operator<< knows it should do its usual bit-shifting behavior.

The same applies for operator>>.

Note that if you're using operator << for both output and left shift, parenthisization is required:

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#include <iostream> #include <bitset> int main() { std::bitset<4> x{ 0b0110 }; std::cout << x << 1 << '\n'; // print value of x (0110), then 1 std::cout << (x << 1) << '\n'; // print x left shifted by 1 (1100) return 0; } |

This prints:

01101 1100

The first line prints the value of x (0110), and then the literal 1. The second line prints the value of x left-shifted by 1 (1100).

We will talk more about operator overloading in a future section, including discussion of how to overload operators for your own purposes.

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.

Flipping 4 bits:

~0100 is 1011

Flipping 8 bits:

~0000 0100 is 1111 1011

In both the 4-bit and 8-bit cases, we start with the same number (binary 0100 is the same as 0000 0100 in the same way that decimal 7 is the same as 07), but we end up with a different result.

We can see this in action in the following program:

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#include <iostream> #include <bitset> int main() { std::cout << std::bitset<4>(~0b0100u) << ' ' << std::bitset<8>(~0b0100u); return 0; } |

This prints:

1011 11111011

Bitwise OR

Bitwise OR (|) works much like its *logical OR* counterpart. However, instead of applying the *OR* to the operands to produce a single result, *bitwise OR* applies to each bit! For example, consider the expression `0b0101 | 0b0110`

.

To do (any) bitwise operations, it is easiest to line the two operands up like this:

0 1 0 1 0 1 1 0

and then apply the operation to each *column* of bits.

If you remember, *logical OR* evaluates to *true (1)* if either the left, right, or both operands are *true (1)*, and *0* otherwise. *Bitwise OR* evaluates to *1* if either the left, right, or both bits are *1*, and *0* otherwise. Consequently, the expression evaluates like this:

0 1 0 1 0 1 1 0 ------- 0 1 1 1

Our result is 0111 binary.

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#include <iostream> #include <bitset> int main() { std::cout << (std::bitset<4>(0b0101) | std::bitset<4>(0b0110)); return 0; } |

This prints:

0111

We can do the same thing to compound OR expressions, such as `0b0111 | 0b011 | 0b0001`

. If any of the bits in a column are *1*, the result of that column is *1*.

0 1 1 1 0 0 1 1 0 0 0 1 -------- 0 1 1 1

Here's code for the above:

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#include <iostream> #include <bitset> int main() { std::cout << (std::bitset<4>(0b0111) | std::bitset<4>(0b0011) | std::bitset<4>(0b0001)); return 0; } |

This prints:

0111

Bitwise AND

Bitwise AND (&) works similarly to the above. *Logical AND* evaluates to true if both the left and right operand evaluate to *true*. *Bitwise AND* evaluates to *true (1)* if both bits in the column are *1*. Consider the expression `0b0101 & 0b0110`

. Lining each of the bits up and applying an AND operation to each column of bits:

0 1 0 1 0 1 1 0 -------- 0 1 0 0

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#include <iostream> #include <bitset> int main() { std::cout << (std::bitset<4>(0b0101) & std::bitset<4>(0b0110)); return 0; } |

This prints:

0100

Similarly, we can do the same thing to compound AND expressions, such as `0b0001 & 0b0011 & 0b0111`

. If all of the bits in a column are 1, the result of that column is 1.

0 0 0 1 0 0 1 1 0 1 1 1 -------- 0 0 0 1

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#include <iostream> #include <bitset> int main() { std::cout << (std::bitset<4>(0b0001) & std::bitset<4>(0b0011) & std::bitset<4>(0b0111)); return 0; } |

This prints:

0001

Bitwise XOR

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 its operands is *true (1)*. If neither or both are true, it evaluates to *0*. Consider the expression `0b0110 ^ 0b0011`

:

0 1 1 0 0 0 1 1 ------- 0 1 0 1

It is also possible to evaluate compound XOR expression column style, such as `0b0001 ^ 0b0011 ^ 0b01111`

. 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 0 0 1 1 0 1 1 1 -------- 0 1 0 1

Bitwise assignment operators

Similar to 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 `x = x >> 1;`

, you can write `x >>= 1;`

.

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#include <bitset> #include <iostream> int main() { std::bitset<4> bits { 0b0100 }; bits >>= 1; std::cout << bits; return 0; } |

This program prints:

0010

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.

In the next lesson, we'll explore how these operators can be used in conjunction with bit masks to facilitate bit manipulation.

Quiz time

Question #1

a) What does 0110 >> 2 evaluate to in binary?

b) What does the following evaluate to in binary: 0011 | 0101?

c) What does the following evaluate to in binary: 0011 & 0101?

d) What does the following evaluate to in binary (0011 | 0101) & 1001?

Question #2

A bitwise rotation is like a bitwise shift, except that any bits shifted off one end are added back to the other end. For example 0b1001 << 1 would be 0b0010, but a left rotate by 1 would result in 0b0011 instead. Implement a function that does a left rotate on a std::bitset<4>. For this one, it's okay to use std::bitset<4>::test() and std::bitset<4>::set().

The following code should execute:

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#include <iostream> #include <bitset> std::bitset<4> rotl(std::bitset<4> bits) { // Your code here } int main() { std::bitset<4> bits1{ 0b0001 }; std::cout << rotl(bits1) << '\n'; std::bitset<4> bits2{ 0b1001 }; std::cout << rotl(bits2) << '\n'; } |

and print the following:

0010 0011

O.3 -- Bit manipulation with bitwise operators and bit masks |

Index |

O.1 -- Bit flags and bit manipulation via std::bitset |

So if

14 >> 1 = 1110 => 0111 = 7 = 14/2^1

16 >> 2 = 10000=>00100 = 4 = 16/2^2

24 >> 3 = 11000=>00011 = 3 = 24/2^3

Then so long as bits aren't lost off the end the bit shifting operators also work to either multiply or divide by 2 raised to the power of the amount of bits shifted by

I assume this is never recommended but just the way that stuff works in any base

While you can multiple or divide by 2 by shifting bits, it isn't recommended that you do this manually. It's better to be clear about your intent, and then let the compiler handle the optimization of this for you.

I came across exactly what we had been learning in this lesson over the week-end, when I looked up how to turn off the tool tip in windows. There was some discussion about it on a forum and someone had came across an entry in the registry, using flags set in hex, exactly like Alex had show us, but the guy had added the caveat that it look awfully complicated and should probably be left alone. I could see straight away that bit 7 needed to be turned off to switch off the tool tips. The number was a 2 byte hex number, and using the calculator that comes with Windows in programmer mode and hex number system, it also shows you the number in binary underneath the hex, so I was able to share with the forum the number in hex that the mask needed to be changed to to turn off the feature and felt very pleased with myself, lol.

Sorry but i think this sentence has a little mistake "If the left operand is an integer type, then operator<< knows it should do its usual bit-shifting behavior."

As an example:

it will output 23 (no shift left occured)

but if you want to do shift left you can do this:

Thank you

The sentence is correct.

In the following code snippet, operator << evaluates from left to right. So this code:

cout<<2 evaluates first, printing 2, and then then cout<<3 evaluates, printing 3. Putting parenthesis changes the precedence, so (2<<3) evaluates first, causing this to do a bit shift and then print the value.

~x - flips all the bits of int x.

Then, is there any option to flip a particular bit of the integer?

Yes, the bitwise exclusive OR operator (operator^) can flip a particular bit.

e.g. myflags ^ 0x10; // flip the second to last bit

hi Alex

how are you doing. hope fine. you are doing a great job.

I have two questions,

a) myflags ^|0x10; you wrote in your reply above, please can you explain, ^|, also which bit was flipped(is it the binary of 0x10)

b) if you want to toggle 2 bit states, why is it - myflags ^= option4 | option5; and not myflags ^= option4 ^ option5;

cheers

1) Typo. I've updated the comment.

2) If you want to toggle 2 bit states, myflags ^= option4 | option5 evaluates as myflags ^= (option4 | option5), where (option4 | option5) turn bit flags 4 and 5 on. Then those are xor'd with myflags. Makes sense logically, right?

myflags ^= option4 ^ option5 will evaluate to myflags ^= 0, because (option4 ^ option5) evaluates to (1 ^ 1) which is 0. Definitely not what you want.

could anyone sove this program for meplzzzzzzzz

Bitwise example, run it and print out the results, reprogram it using OOP (c++)

concepts

#include

main ()

{

unsigned char x = '11';

unsigned char y = '27';

cout << "x = " << oct << short(x) << '\n';

cout << "y = " << oct << short(y) << '\n';

cout << "~x = " << oct << (short)(~x) << '\n';

cout << "x & y = " << oct << (short)(x & y) << '\n';

cout << "x | y = " << oct << (short)(x | y) << '\n';

cout << "x ^ y = " << oct << (short)(x ^ y) << '\n';

cout << "x << 2 = " << oct << (short)(x << 2) << '\n';

cout <> 2 = " << oct <> 2) << '\n';

pls reply to the questions

Which of the following statements will display "Problem!" if bit 2 of flags is a '0'?

Answer if ((flags & 0x04) == 0x04) cout << "Problem!";

if ((flags & 0x04) != 0x04) cout << "Problem!";

if ((flags ^ 0x04) == 0x04) cout << "Problem!";

if ((flags ^ 0x04) != 0x04) cout << "Problem!";

2 points

To turn bit 3 of a variable to 0. the correct way is to:

Answer set the mask to 0xEF and peform the bitwise AND operation with the variable.

set the mask to 0xF7 and peform the bitwise AND operation with the variable.

set the mask to 0xEF and peform the bitwise OR operation with the variable.

set the mask to 0xF7 and peform the bitwise OR operation with the variable.