O.3 — Bit manipulation with bitwise operators and bit masks

In the previous lesson on bitwise operators (O.2 -- Bitwise operators), we discussed how the various bitwise operators apply logical operators to each bit within the operands. Now that we understand how they function, let’s take a look at how they’re more commonly used.

Bit masks

In order to manipulate individual bits (e.g. turn them on or off), we need some way to identify the specific bits we want to manipulate. Unfortunately, the bitwise operators don’t know how to work with bit positions. Instead they work with bit masks.

A bit mask is a predefined set of bits that is used to select which specific bits will be modified by subsequent operations.

Consider a real-life case where you want to paint a window frame. If you’re not careful, you risk painting not only the window frame, but also the glass itself. You might buy some masking tape and apply it to the glass and any other parts you don’t want painted. Then when you paint, the masking tape blocks the paint from reaching anything you don’t want painted. In the end, only the non-masked parts (the parts you want painted) get painted.

A bit mask essentially performs the same function for bits -- the bit mask blocks the bitwise operators from touching bits we don’t want modified, and allows access to the ones we do want modified.

Let’s first explore how to define some simple bit masks, and then we’ll show you how to use them.

Defining bit masks in C++14

The simplest set of bit masks is to define one bit mask for each bit position. We use 0s to mask out the bits we don’t care about, and 1s to denote the bits we want modified.

Although bit masks can be literals, they’re often defined as symbolic constants so they can be given a meaningful name and easily reused.

Because C++14 supports binary literals, defining these bit masks is easy:

Now we have a set of symbolic constants that represents each bit position. We can use these to manipulate the bits (which we’ll show how to do in just a moment).

Defining bit masks in C++11 or earlier

Because C++11 doesn’t support binary literals, we have to use other methods to set the symbolic constants. There are two good methods for doing this. Less comprehensible, but more common, is to use hexadecimal. If you need a refresher on hexadecimal, please revisit lesson 4.12 -- Literals.

This can be a little hard to read. One way to make it easier is to use the left-shift operator to shift a bit into the proper location:

Testing a bit (to see if it is on or off)

Now that we have a set of bit masks, we can use these in conjunction with a bit flag variable to manipulate our bit flags.

To determine if a bit is on or off, we use bitwise AND in conjunction with the bit mask for the appropriate bit:

This prints:

bit 0 is on
bit 1 is off

Setting a bit

To set (turn on) a bit, we use bitwise OR equals (operator |=) in conjunction with the bit mask for the appropriate bit:

This prints:

bit 1 is off
bit 1 is on

We can also turn on multiple bits at the same time using Bitwise OR:

Resetting a bit

To clear a bit (turn off), we use Bitwise AND and Bitwise NOT together:

This prints:

bit 2 is on
bit 2 is off

We can turn off multiple bits at the same time:

Flipping a bit

To toggle a bit state, we use Bitwise XOR:

This prints:

bit 2 is on
bit 2 is off
bit 2 is on

We can flip multiple bits simultaneously:

Bit masks and std::bitset

std::bitset supports the full set of bitwise operators. So even though it’s easier to use the functions (test, set, reset, and flip) to modify individual bits, you can use bitwise operators and bit masks if you want.

Why would you want to? The functions only allow you to modify individual bits. The bitwise operators allow you to modify multiple bits at once.

This prints:

bit 1 is off
bit 2 is on
bit 1 is on
bit 2 is off
bit 1 is on
bit 2 is on
bit 1 is off
bit 2 is off

Making bit masks meaningful

Naming our bit masks “mask1” or “mask2” tells us what bit is being manipulated, but doesn’t give us any indication of what that bit flag is actually being used for.

A best practice is to give your bit masks useful names as a way to document the meaning of your bit flags. Here’s an example from a game we might write:

Here’s the same example implemented using std::bitset:

Two notes here: First, std::bitset doesn’t have a nice function that allows you to query bits using a bit mask. So if you want to use bit masks rather than positional indexes, you’ll have to use Bitwise AND to query bits. Second, we make use of the any() function, which returns true if any bits are set, and false otherwise to see if the bit we queried remains on or off.

When are bit flags most useful?

Astute readers may note that the above examples don’t actually save any memory. 8 booleans would normally take 8 bytes. But the above examples use 9 bytes (8 bytes to define the bit masks, and 1 bytes for the flag variable)!

Bit flags make the most sense when you have many identical flag variables. For example, in the example above, imagine that instead of having one person (me), you had 100. If you used 8 Booleans per person (one for each possible state), you’d use 800 bytes of memory. With bit flags, you’d use 8 bytes for the bit masks, and 100 bytes for the bit flag variables, for a total of 108 bytes of memory -- approximately 8 times less memory.

For most programs, the amount of memory using bit flags saved is not worth the added complexity. But in programs where there are tens of thousands or even millions of similar objects, using bit flags can reduce memory use substantially. It’s a useful optimization to have in your toolkit if you need it.

There’s another case where bit flags and bit masks can make sense. Imagine you had a function that could take any combination of 32 different options. One way to write that function would be to use 32 individual Boolean parameters:

Hopefully you’d give your parameters more descriptive names, but the point here is to show you how obnoxiously long the parameter list is.

Then when you wanted to call the function with options 10 and 32 set to true, you’d have to do so like this:

This is ridiculously difficult to read (is that option 9, 10, or 11 that’s set to true?), and also means you have to remember which parameters corresponds to which option (is setting the “edit flag” the 9th, 10th, or 11th parameter?) It may also not be very performant, as every function call has to copy 32 booleans from the caller to the function.

Instead, if you defined the function using bit flags like this:

Then you could use bit flags to pass in only the options you wanted:

Not only is this much more readable, it’s likely to be more performant as well, since it only involves 2 operations (one Bitwise OR and one parameter copy).

This is one of the reasons OpenGL, a well regarded 3d graphic library, opted to use bit flag parameters instead of many consecutive Boolean parameters.

Here’s a sample function call from OpenGL:

GL_COLOR_BUFFER_BIT and GL_DEPTH_BUFFER_BIT are bit masks defined as follows (in gl2.h):

Bit masks involving multiple bits

Although bit masks often are used to select a single bit, they can also be used to select multiple bits. Lets take a look at a slightly more complicated example where we do this.

Color display devices such as TVs and monitors are composed of millions of pixels, each of which can display a dot of color. The dot of color is composed from three beams of light: one red, one green, and one blue (RGB). By varying the intensity of the colors, any color on the color spectrum can be made. Typically, the amount of R, G, and B for a given pixel is represented by an 8-bit unsigned integer. For example, a red pixel would have R=255, G=0, B=0. A purple pixel would have R=255, G=0, B=255. A medium-grey pixel would have R=127, G=127, B=127.

When assigning color values to a pixel, in addition to R, G, and B, a 4th value called A is often used. “A” stands for “alpha”, and it controls how transparent the color is. If A=0, the color is fully transparent. If A=255, the color is opaque.

R, G, B, and A are normally stored as a single 32-bit integer, with 8 bits used for each component:

32-bit RGBA value
bits 31-24 bits 23-16 bits 15-8 bits 7-0
red green blue alpha

The following program asks the user to enter a 32-bit hexadecimal value, and then extracts the 8-bit color values for R, G, B, and A.

This produces the output:

Enter a 32-bit RGBA color value in hexadecimal (e.g. FF7F3300): FF7F3300
Your color contains:
ff red
7f green
33 blue
0 alpha

In the above program, we use a bitwise AND to query the set of 8 bits we’re interested in, and then we right shift them into an 8-bit value so we can print them back as hex values.


Summarizing how to set, clear, toggle, and query bit flags:

To query bit states, we use bitwise AND:

To set bits (turn on), we use bitwise OR:

To clear bits (turn off), we use bitwise AND with bitwise NOT:

To flip bit states, we use bitwise XOR:

Quiz time

Question #1

Given the following program:

a) Write a line of code to set the article as viewed.

Show Solution

b) Write a line of code to check if the article was deleted.

Show Solution

c) Write a line of code to clear the article as a favorite.

Show Solution

1d) Extra credit: why are the following two lines identical?

Show Solution

O.4 -- Converting between binary and decimal
O.2 -- Bitwise operators

14 comments to O.3 — Bit manipulation with bitwise operators and bit masks

  • Chayim

    You said:
    "We use 0s to mask out the bits we don’t care about, and 1s to denote the bits we want modified."
    Where do you use 0s in this lesson to "mask out" bits and 1s to "denote bits"? And how does 0s and 1s mask out or in bits?

  • Chayim

    You said that "cstdint" probably stands for "standard integer" that’s what I knew but what I don’t know is what does the "c" stand for?

  • Chayim

    Cannot reply to comments, also there’s no timer to edit a comment.

  • Chayim

    How does flags "|= mask1" turn on a bit?

  • nascardriver

    "standard integer" probably.
    There are many `flags` on this page. The `flags` in "Summary" assumes there is a `flags` bitset or integer, it doesn't matter.

  • Chayim

    What does "cstdint" stand for?
    And where was flags defined?

  • Rishi

    We use hexadecimal values only because inputting binary values for all the 32 bits is time elapsing. Right?

  • Rishi

    I have a doubt. Maybe it's silly. In the program which asks the user to enter a 32-bit hexadecimal value, and then extracts the 8-bit color values for R, G, B, and A, we assign 0xFF000000 to the red extractor. When the user enters a value, he/she enters a value of 8 bit long. I don't understand this. We are defining 32bit long value and inputting and operating using 8 bit long values.

    • nascardriver

      The user enters 0xFF000000, which is a 32bit value. The program splits this 32bit value into 4 8bit values.

      • Rishi

        You mean, the limit for the data type is 32 bits? How's FF000000 32 bit value??? It's 8 bits isn't it?

        • nascardriver

          FF is the largest 8bit value. Anything beyond FF doesn't fit into 8 bits.
          FF000000 is a 32bit value.
          Maybe you're being confused by the trailing zeros. hex(FF000000) is dec(4278190080).

          • Rishi

            Oh yeah yeah. It's all clear now. Thank you for helping me understand. And I am amazed by the way you keep this website updated and answer every questions asked. May God bless you and your team. Keep helping others✌

  • Chayim

    Masktape is not the appropriate example to bitmask because it’s the opposite, masking against paint is for the cause to exclude the masked part but with bitmask it’s the cause of using only that bit and excluding all other bits.

  • mike

    when i finshed the last lesson i did some functions to do set,reset,test,flip
    before i know about bit masks
    i want to know are my functions is usable or not ?
    sorry for bad english

    here is my code :

    #include <iostream>
    #include <bitset>
    #define typeOfNumber unsigned char // type of bitflags(should be unisgned integer)
    constexpr char numberOfBits{ 8 }; // should identical with the type above

    // to give me the position (i didn't now about the bit masks yet)
    int expontaion(int value, int exponent) {
        int theResult{ 1 };
        for (;exponent > 0;exponent--) {
            theResult *= value;
        return theResult;
    // set bit to 1
    void setBit(typeOfNumber & bitFlags, typeOfNumber position)
        position = expontaion(2, position);
        bitFlags = bitFlags | position;
    // set bit to 0
    void reSetBit(typeOfNumber& bitFlags, typeOfNumber position)
        position = expontaion(2, position);
        if ((bitFlags & position) == position)
            bitFlags ^=  position;
    // if bit is 0 set it to 1 (vice versa)
    void flipBit(typeOfNumber& bitFlags, unsigned typeOfNumber position)
        position = expontaion(2, position);
        bitFlags = bitFlags ^ position;
    //test bit
    typeOfNumber getBit(typeOfNumber  bitFlags, typeOfNumber position)
        bitFlags = bitFlags << (numberOfBits - 1) - position;
        bitFlags = bitFlags >>( numberOfBits - 1);
        return bitFlags;

    int main() {
        unsigned char infoBitFlags{0b1010'1010};
        setBit(infoBitFlags, 0);                            // set position 1 in infoBitFlags to 1
        std::cout << std::bitset<8>{infoBitFlags} << '\n';    //this will print 1010'1011
        return 0;

  • Imagine

    I don't know what it particularly is but, it feels like I didn't understand something. I even read these chapters a couple of times, thoroughly too, but it still feels like I am not completely sure about it. It might have to do with bits and bytes and values, I don't know. Due to this I have not been able to proceed further...

    • Ryan

      Try subbing in the values and test it out. For example if we test if mask1 is turned on. We can do (flag & mask1)

      Subbing in we get:
      0000'0101 &

      Remember that for '&', both bytes have to be true (1). Since all the whole byte is false(they are all 0), we can say that mask1 is turned off.

      You may notice the bytes that are turned on have at least 1 true bit for flag & masknumber.

      So when we are turning on a bit, we use OR '|', which means at least one or more bit have to be true(1). This guarentees that at least one of the bytes will have a 1, for example:

      flag |= mask1 =

      0000'0101 OR
      0000'0111    As there is more than one true bit, mask1 is turned on.

      Now lets see what we get when we turn off a bit.
      From the first example above, we are told that mask2 = 0000'0100 is turned on.
      We can turn it off with flag &= ~mask2. Let's see how it works:

      ~mask2 = 1111'1011.
      flag = flag & ~mask2 is:

      0000'0101 AND

      Since all the bits are 0, we say that mask2 is turned off now.

      I might be completely wrong, but this is the way I understood this chapter. Please correct me if I'm wrong in anything.

  • James

    Hi there, awesome series of lessons you've all got going here, really appreciate the effort that goes into maintaining a site like this, kudos!

    I have a question regarding the final example before the summary. As far as I can tell, my code is identical to that provided in the example;

    However I keep receiving a compiler warning telling me my green RGBA component variable in main() is being initialized with a 'possible loss of data', implying that the data lost are the zeros resulting from the 16 right bitshifts of (green & greenMask), even though the other component variables all compile with no issue at all. I had to disable treating warnings as errors in order for my program to compile. Suffice to say that it works fine, so I'm just wondering why I am receiving the aforementioned warning? Thanks in advance.

  • Jack

    constexpr std::uint_fast8_t isHungry{    1 << 0 }; // 0000 0001
        constexpr std::uint_fast8_t isSad{        1 << 1 }; // 0000 0010
        constexpr std::uint_fast8_t isMad{        1 << 2 }; // 0000 0100
        constexpr std::uint_fast8_t isHappy{    1 << 3 }; // 0000 1000
        constexpr std::uint_fast8_t isLaughing{ 1 << 4 }; // 0001 0000
        constexpr std::uint_fast8_t isAsleep{    1 << 5 }; // 0010 0000
        constexpr std::uint_fast8_t isDead{        1 << 6 }; // 0100 0000
        constexpr std::uint_fast8_t isCrying{    1 << 7 }; // 1000 0000

    Why are we not using binary numbers here, but a left shift? And how the value of 1 is evaluated.

  • Obito

    #include <iostream>
    #include <cstdint>
    #include <bitset>

    int main()
        constexpr std::uint_fast8_t option1{ 0b0000'0000 };
        constexpr std::uint_fast8_t option2{ 0b0000'0001 };
        constexpr std::uint_fast8_t option3{ 0b0000'0010 };
        constexpr std::uint_fast8_t option4{ 0b0000'0100 };
        constexpr std::uint_fast8_t option5{ 0b0000'1000 };
        constexpr std::uint_fast8_t option6{ 0b0001'0000 };
        constexpr std::uint_fast8_t option7{ 0b0010'0000 };
        constexpr std::uint_fast8_t option8{ 0b0100'0000 };
        std::uint_fast8_t bitflag{ };

    /* constexpr std::bitset<8> option1{ 0b0000'0000};
        constexpr std::bitset<8> option2{ 0b0000'0001 };
        constexpr std::bitset<8> option3{ 0b0000'0010 };
        constexpr std::bitset<8> option4{ 0b0000'0100 };
        constexpr std::bitset<8> option5{ 0b0000'1000 };
        constexpr std::bitset<8> option6{ 0b0001'0000 };
        constexpr std::bitset<8> option7{ 0b0010'0000 };
        constexpr std::bitset<8> option8{ 0b0100'0000 };
        std::bitset<8> bitflag{};*/ (this one can print out binary number)

        bitflag |= option7;
        std::cout << bitflag;
        return 0;
    When i had used uint_fast8_t, it had printed nothing. But, if i had used bitset, it had printed out the binary number. Can i known why it had been like that, what is my problem make it can not print out if i used uint_fast8_t (for the information, the compiler does not show any warning or error)

    • nascardriver

      `std::uint_fast8_t` might be an `unsigned char`. `std::cout <<` treats `char` as characters. Add a cast to `int`.

      • Obito

        when i used std::uint_fast16_t, it can print out 32. What should i do if i want to print out on binary (0010'0000), not on decimal (32)? If i used static_cast<int>(bitflag), it will going to print out decimal too. Should i make a function for let it print out binary number?

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