O.2 — Bitwise operators

The bitwise operators

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 prefix 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 operand 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:

This prints:


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


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:

This program prints:


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 is an integral 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, parenthesization is required:

This prints:


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:

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 OR
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 OR
0 1 1 0
0 1 1 1

Our result is 0111 binary.

This prints:


We can do the same thing to compound OR expressions, such as 0b0111 | 0b0011 | 0b0001. If any of the bits in a column are 1, the result of that column is 1.

0 1 1 1 OR
0 0 1 1 OR
0 0 0 1
0 1 1 1

Here's code for the above:

This prints:


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 AND
0 1 1 0
0 1 0 0

This prints:


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 AND
0 0 1 1 AND
0 1 1 1
0 0 0 1

This prints:


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 XOR
0 0 1 1
0 1 0 1

It is also possible to evaluate compound XOR expression column style, such as 0b0001 ^ 0b0011 ^ 0b0111. 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 XOR
0 0 1 1 XOR
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;.

This program prints:



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?

Show Solution

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

Show Solution

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

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

Show Solution

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 test() and set().

The following code should execute:

and print the following:


Show Solution

Question #3

Extra credit: Redo quiz #2 but don't use the test and set functions.

Show Solution

O.3 -- Bit manipulation with bitwise operators and bit masks
O.1 -- Bit flags and bit manipulation via std::bitset

231 comments to O.2 — Bitwise operators

  • Dudz

    My answer for quiz#2 seems fine.

    but this answer to quiz 3 seems a little ugly :)

  • Michael Okolo

    I dont know when I will stop over complicating things. This is what I came up with for Question 2(rotl) and 3(rrotl).

  • Surendar Raj G

    I did not understand how quiz 3 worked .

    can any one give a detailed explanation over it.

    • nascardriver

      I explained it here
      If that doesn't help, please leave a reply.

      • Dudz

        Hi nascardriver,
        After posting my answer for quiz 2 and 3, I tried this

        but it yields this wrong answer


        When I calculate it manually, it really is 1010 given bits1 and bits2 values.
        For bits1: 0001
        return (bits<<1) | (bits<<3);
                0001<<1       0001<<3
                0010          | 1000
        it returns 1010.

        For bits2: 1001
        return (bits<<1) | (bits<<3);
                1001<<1      1001<<3
                0010          | 1000
        it also returns 1010. Leftmost bit was gone after shifting to the left and cannot transfer to bit(0)

  • RJ

    This was my first try on Question 3




  • Am

    Initially I didn't think to use the OR operand to make a couple of moves and ended up making something overly complicated. Hope this gives you a good laugh.

    This is the optimized version after I checked the solution:

  • Ryan

    In paragraph 'What!? Aren't operator<< and operator>> used for input and output?': "If the left operand some other type". Missing an 'is'!

  • Sam

    Grammatical error. "Note that if you're using operator << for both output and left shift, parenthisization is required:". Change "parenthisization" to parenthesization.

  • kavin

    Can i do like this too?

    std::bitset<4> rotl(std::bitset<4> bits)
        if (bits.test(3))
            bits=bits << 1; //0010
            return bits;

            bits=bits <<1;
        return bits;

    int main()
        std::bitset<4> bits1{ 0b0001 };
        std::cout << rotl(bits1) << '\n';

        std::bitset<4> bits2{ 0b1001 };
        std::cout << rotl(bits2) << '\n';

        return 0;


    I was interested in why this code doesn't work for question 2:

    (12, 10): error C2059: syntax error: '<<='

    Also getting unexpected results for this:

    Results I get:

    Only first 2 operations give expected results

    • nascardriver

      `<<=` is a binary operator, it needs two operands, you only gave one. `<<=` modifies the left operand, but you don't need that. I assume you want

      Also note that `test` returns a `bool`, not an integer.

      Those aren't binary numbers, they're missing the `0b` prefix.

      • XpORPID

        Oh, even after looking through the first code a couple of times I was still looking at the line 8 when the errors was referring to line 12.
        Stupid mistakes by me..

        Thank you, nascardriver!

  • knight

    Can someone explain and analyze the question 2 answer?

    • nascardriver

      • knight

        Thanks nascardriver. But I am confused about the:


        What is 3 meaning, how leftmost bit 1 become rightmost bit 1.

        • nascardriver

          Bits are numbered right to left, starting at 0.

          `bits.test(3)` returns true if the leftmost bit is 1, otherwise it returns false.
          `bits.set(0)` sets the rightmost bit to 1. We only do this if the leftmost bit was 1 before the shift.

  • Ged

    You are using "u" at the end of the bit numbering. Is it because we are using ~ and it requires "u" (it stands for unsigned type) or something else? Cause other examples don't have it.

    • nascardriver

      The first bit of signed integers is used as the sign bit. If it's 1, the integer is considered to be negative, otherwise it's positive. `0b0100`'s first bit is 0, so `~0b0100`'s first bit is 1. That means that the number is negative, but `std::bitset` doesn't work with negative numbers. The bits of unsigned integers don't have a special meaning, because all unsigned integers are non-negative. `0b0100u` is positive and so is `~0b0100u`.

  • Vlad dude

    Bitwise XOR explanation,second paragraph,there's an extra bit inside the third value.
    Thought I wasn't understanding something at first lol

  • Mike

    I just completed Ch.O after finishing up Ch.5 before finally noticing the notification regarding the "disjointed lesson numbers". I know it said it was optional, and I was thinking the whole time, these lessons seem out of place.

    So where exactly should Ch.O be?

    • Alex

      The lessons should be completed in the order they are listed on the front page of the site, with O slotting in between 5 and 6. Perhaps I should rename O as lesson 6, and renumber 6, so the intended ordering is clearer. I'll think about doing so when I roll out the update to the current chapter 6.

  • D

    Concerning best practices:
    return 0; statement missing in Question #2.

  • MorTaZz

    hi, i know you are a really busy person but would you take a look at this code and tell me why doesnt it work?

  • Aakash

    Typo Alert!!
    "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.",

    There should be 0b0011 instead of 0b011.

  • Chris

    The solution to question 2 seems kind of lengthy.  After all, you just need to return (bits<<1)|(bits>>3) from the rotl function.

    • Alex

      You're right. I added this as an extra credit question/solution and gave you a tip o' the hat. Thanks!

      • giang

        Hi, I tried with the real binary & your solution work really great despite its simplicity. But I still can't find or understand the rule behind this solution. Can you explain more clearly??

        • nascardriver

          Let's say we have a number

          where A, B, C, and D are bits, eg. 0110

          First, we perform the left shift (`bits << 1`)

          We lost the A and got a 0 on the right. Now we perform the right shift (`bits >> 3`) (On the original ABCD)

          We lost everything apart from the A, which is now the least significant bit.

          All we have to do now is combine the two results

  • Mathew

    Sorry if it seems like I'm being nitpicky here.

    On the first 'Author notes', 0b is mentioned as a suffix, and
    it took me a while to understand that it was meant to be a prefix.

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

    Thanks for the tutorial!

  • avidlearner

    Isn't there a mistake in the solution to Quiz question 1) d)?

    You're doing an AND operation instead of an OR.

  • mansbota

    Shouldn't this print -126, since we're dealing with signed char? I get -2.
    EDIT: nvm, just realized that it's getting value from two complement method. Using signed and absolute value, it appears that char's range is from -127 to 127 (255), while unsigned the range is from 0 to 255 (256)

    Also, why doesn't this compile:

  • Randle

    (EDIT) I just looked at the next lesson and realized this is probably more simply done by using an unsigned char, but this still works. :-)

    If anyone wants something to try in terminal to see how bitwise operations work on actual c++ datatypes, we can define a short with a binary value (even though they are saved as ints) and follow along with operations.

  • NooneAtAll

    What happens if we bitwise AND between variables of different sizes?


    Does it start from the beginning of both?
    Does it loop through the smaller one?

    • The smaller types are promoted to the larger type.
      @oneByte -> int
      @manyBytes -> long long

      If both types are smaller than an int, they're promoted to an int.

  • rohit

    int is of 32 bits normally.
    so why ~3 is giving ans -4 instead of 4294967292. all 32 bits will be flipped if its is an integer?

  • Nathan

    Did the site go down yesterday?

  • nullptr

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

    Shouldn't it be overload instead of overrride?

  • Ajalle Perfej

    I'd like to ask your comments on some uses of bitwise operators I found on the internet (link available upon request).

    One of them is this form of the postfix initializer and incrementer:

    To be fair to the obviously experienced programmer who provides this example for Javascript rather than C++, he calls this use of bitwise operators "evil" because it's compiler abuse just in case you forgot to initialize your incrementer variable.

    Another use just punched me in the face, and it's for rounding a float into an integer:

    I figure that shortcut is probably most useful in environments such as embedded programming, but it's my first exposure to memory conservation and I find it stimulating. With regard to C++ or C can you point me to some resources on programming for low-memory environments? I'm happy with just one link where I can learn the basics.

    Finally, here's one of my own:

    In a very old computer RPG, there were four basic classes: Cleric, Fighter, Magic-User, and Thief. You had a party of six characters, although you could try to win the game with fewer than six. Some races of characters could be two or three classes at the same time. Well, in order to have a reasonable chance at winning the game, you needed 10 character classes among your six characters. There's a way to select those randomly using bitwise logic:

    Each character starts with a class value of 0.

    Cleric: character | 1
    Fighter: character | 2
    Magic-User: character | 4
    Thief: character | 8

    You can code a function that pseudorandomly flips the bits of your six characters until you have your 10 starting character classes, and build your pseudorandom party of six characters that way. It can be really fun just to go with whatever the pseudorandom generator gives you, and there's no way within the game to make pseudorandom character class choices, so this is one way. Neat, eh?

    • Hi Ajalle!

      > One of them is this form of the postfix initializer and incrementer [...]
      Nope. If @incrementer is uninitialized and not 0, the line after won't magically turn it into a 1. The example works in JS, because uninitialized values in JS are have the value 'undefined'. '~undefined' is -1, so '-~undefined' is 1. Uninitialized variables in C++ have an unspecified value. Always initialize your variables!

      > Another use just punched me in the face
      Again, this works in JS, but not in C++. floats don't have an @operator|.

  • Haroon

    Is XOR the only bitwise operator that doesn't cause data loss?
    Operation reversal can evaluate the original value?

    If, yes, I think this should be noted here. If, no, please correct me.

  • Aleksandr

    "Note: Many people find this lesson challenging." And yet you explained the concepts so clearly that it became trivial to understand :) Thanks!

  • Eric

    I’d like to add something my high school math teacher taught me back in 1975.

    OR turns bits *on*.
    AND turns bits *off*.
    XOR *toggles* bits.

    (Edit: Sorry Alex, I see now you say this exact thing in the next chapter!)

  • Vu Dang


    "With an 8-bit value, 3 << 3 would be 24 because the 16-bit wouldn't be shifted off the end of the binary number.", at the "8-bit" and "16-bit"

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