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, at least for application programming, it is probably 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, games where bit manipulation tricks can be used for extra speed, or embedded programming, where memory is still limited). 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 |

Bit manipulation is one of the few cases where you should unambiguously use unsigned integer data types. This is because C++ does not guarantee how signed integers are stored, nor how some bitwise operators apply to signed variables.

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).

*Rule: When dealing with bit operators, use unsigned integers.*

**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, and the right operator is an integer number of bits to shift by. So when we say `3 << 1`

, we are saying “shift the bits in the literal 3 left by 1 place”.

For example, consider the number 3, which is binary 0011:

3 = 0011

3 << 1 = 0110 = 6

3 << 2 = 1100 = 12

3 << 3 = 1000 = 8

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.

12 = 1100

12 >> 1 = 0110 = 6

12 >> 2 = 0011 = 3

12 >> 3 = 0001 = 1

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

Although our examples above involve shifting literals, you can shift variables as well:

1 2 |
unsigned int x = 4; x = x << 1; // x will be 8 |

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

**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 cout to output text. Consider the following program:

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

This program prints:

8

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 **overridden** (replaced) the default behavior of the << operator and given it a new meaning (to output text to the console). 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 overrode to do output. If the left operand is an integer type, then operator<< knows it should do its usual bit-shifting behavior.

We will talk more about operator overloading in a future section, including discussion of how to override 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!

Assuming 4 bits:

4 = 0100

~4 = 1011 = 11

Assuming 8 bits:

4 = 0000 0100

~4 = 1111 1011 = 251

**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 `x = x << 1;`

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

.

**Bit flags**

As noted above, using a whole byte to store a boolean leaves 7 of the 8 bits unused. While this generally is fine, in storage-intensive cases where we have lots of related boolean options, it can be useful to "pack" 8 individual booleans into a single byte for storage efficiency purposes. These individual bits are called **flags**. Since the bits can't be accessed directly, we have to use the bitwise operators to set, unset, or query them.

For example:

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// Define 8 separate flags (these can represent whatever you want) // Note: in C++11, better to use "uint8_t" instead of "unsigned char" unsigned char option1 = 0x01; // hex for 0000 0001 unsigned char option2 = 0x02; // hex for 0000 0010 unsigned char option3 = 0x04; // hex for 0000 0100 unsigned char option4 = 0x08; // hex for 0000 1000 unsigned char option5 = 0x10; // hex for 0001 0000 unsigned char option6 = 0x20; // hex for 0010 0000 unsigned char option7 = 0x40; // hex for 0100 0000 unsigned char option8 = 0x80; // hex for 1000 0000 unsigned char myflags; // byte-size value to hold some combination of the above 8 options |

To turn a bit on, we use bitwise OR:

1 2 |
myflags |= option4; // turn option 4 on. myflags |= option4 | option5; // turn options 4 and 5 on. |

To turn a bit off, we use bitwise AND with an inverse bit pattern:

1 2 |
myflags &= ~option4; // turn option 4 off myflags &= ~option4 & ~option5; // turn option4 and 5 off. |

To toggle a bit state, we use bitwise XOR:

1 2 |
myflags ^= option4; // flip option4 from on to off, or vice versa myflags ^= option4 | option5; // flip options 4 and 5 |

To query a bit state, we use bitwise AND:

1 |
if (myflags & option4) ... // if option4 is set, do something |

As a real-life example, in OpenGL (a 3d graphics library), some functions take one or more bit flags as a parameter:

1 |
glClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT); // clear the color and the depth buffer |

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

1 2 3 |
#define GL_DEPTH_BUFFER_BIT 0x00000100 #define GL_STENCIL_BUFFER_BIT 0x00000400 #define GL_COLOR_BUFFER_BIT 0x00004000 |

**An RGBA color example**

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" standard for "alpha", and it controls how transparent the color is. If A=0, the color is opaque. If A=255, the color is fully transparent.

R, G, B, and A are normally stored as a 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 |

RRRRRRRR | GGGGGGGG | BBBBBBBB | AAAAAAAA |

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. If you need a reminder about what hexadecimal numbers are, see section 2.8 -- literals.

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#include <iostream> int main() { // Note: in C++11, it's better to use uint8_t than "unsigned char" using namespace std; cout << "Enter a 32-bit RGBA color value in hexadecimal (e.g. FF7F3300): "; unsigned int pixel; cin >> hex >> pixel; // std::hex allows us to read in a hex value // use bitwise AND to isolate red pixels, then right shift the value into the range 0-255 unsigned char red = (pixel & 0xFF000000) >> 24; unsigned char green = (pixel & 0x00FF0000) >> 16; unsigned char blue = (pixel & 0x0000FF00) >> 8; unsigned char alpha = pixel & 0x000000FF; cout << "Your color contains:\n"; cout << static_cast<int>(red) << " of 255 red\n"; cout << static_cast<int>(green) << " of 255 green\n"; cout << static_cast<int>(blue) << " of 255 blue\n"; cout << static_cast<int>(alpha) << " of 255 alpha\n"; } |

This produces the output:

Enter a 32-bit RGBA color value in hexadecimal (e.g. FF7F3300): FF7F3300 Your color contains: 255 of 255 red 127 of 255 green 51 of 255 blue 0 of 255 alpha

Note that different machine architectures may store bytes in a different order. RGBA can be stored as ARGB or BGRA depending on the architecture.

**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.

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

To turn a bit on, we use bitwise OR:

1 2 |
myflags |= option4; // turn option 4 on. myflags |= option4 | option5; // turn options 4 and 5 on. |

To turn a bit off, we use bitwise AND with an inverse bit pattern:

1 2 |
myflags &= ~option4; // turn option 4 off myflags &= ~option4 & ~option5; // turn option4 and 5 off. |

To toggle a bit state, we use bitwise XOR:

1 2 |
myflags ^= option4; // flip option4 from on to off, or vice versa myflags ^= option4 | option5; // flip options 4 and 5 |

To query a bit state, we use bitwise AND:

1 |
if (myflags & option4) ... // if option4 is set, do something |

**Quiz**

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

2) What does 5 | 12 evaluate to in decimal?

3) What does 5 & 12 evaluate to in decimal?

4) What does 5 ^ 12 evaluate to in decimal?

5) Given the following program:

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int main() { unsigned char option_viewed = 0x01; unsigned char option_edited = 0x02; unsigned char option_favorited = 0x04; unsigned char option_shared = 0x08; unsigned char option_deleted = 0x80; unsigned char myArticleFlags; } |

5a) Write a line of code to set the article as viewed.

5b) Write a line of code to check if the article was deleted.

5c) Write a line of code to unset the article as a favorite.

**Quiz answers**

5a) Show Solution

5b) Show Solution

5c) Show Solution

3.x — Comprehensive Quiz |

Index |

3.7 — Converting between binary and decimal |

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…

to miroslav, jeff & sunilla,

there’s a lot of use for bitwise operation.

there might be time when you will have to deal with data represented in binary or some other kind of number base other than decimal or words (as in ASCII)

example 1,

you want to write a software that would evaluate poker hands.

you could encode the hands in binary form.

every suite has 13 cards, there are 4 suites.

assign every suite 2 bytes, which give 16 bits, of which we’ll use 13bits.

each of that bit represent one of the card from A, K, Q, J, 10, 9….. (from bit12 to bit10).

to represent all 4 suites, there are 8 bytes… (32 bit)

if you represent each card in ASCII, like 7d for Seven Diamonds, it takes 2 bytes (1 byte for each ASCII character).

if you want to pass information of 10 cards, it takes you 10 x 2 bytes = 20 bytes.

where as if you use binary information it still gonna take 8 bytes.

plus, if you represent it in ASCII, you’d have to do loopings to sort the hands in order from the highest value to the lowest value. if you do it in binary, you dont have to worry about the order, it’s already in order.

You could also do bitwise operation on your 8-bytes representation of the cards to do comparison, or check if certain card is there or if it’s a flush…. and many more.

It’s an invaluable tool when you need to work smart instead of work hard.

another use is when you decide to take up microcontroller as a hobby. Like Arduino for example, where memory is still a precious commodity and speed is important.

when doing programming you will deal with logic a lot of times…..

sometime your IF statement is gonna be a very long one that it’s hard to read, all entangled in nested parenthesis.

encoding them into binary representation can make it a lot easier.

Maybe if you have electronics background or at least some appreciation in digital logic design, it help you to appreciate bitwise operation more.

correction

correction : each of that bit represent one of the card from A, K, Q, J, 10, 9….. (from bit12 to

bit0).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.

You can’t. The original value is lost as the bits are shifted from the end of the value. Bits are memory-less. They don’t remember previous states. You could, however, use a binary representation to refer to a specific location or value, and use bit shifting to point to those different values.

Binary manipulation allows a fast way to store and manipulate states, as well as performing certain mathematical operations very quickly. Mainly, however, manipulating bits is useful when you are performing a great deal of calculations, where you need a high level of performance, and you’ll find a use for all the bitwise operators here.

Another example; say you were storing the states of a computer opponent in a video game, and each bit referred to a particular action it could take. 1 for on, 0 for off. If each bit represented a particular action, then 5|6 would result in 0 1 0 0, which would refer to only one action the computer A.I could take, where 5 is its current state, and 6 is the next state it finds itself in. This might seem like an obtuse example, but finite state machines are very commonly used in game A.I.

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..

I’ve really enjoyed the tutorial so far, mainly because of the real life examples we’ve been given along the way.

The brief elucidation at the top of this page is great but could you go a bit further and tell us why it’s useful to be able to turn an 8bit 4 into a 251?! What’s the practical use of these operators?

Thanks in advance and thanks for an already brilliant tutorial.

M

I am a college instructor, and have taught Discrete Mathematics using C++, and have also taught video game programming using C++. I’m going through the learncpp website to pick up teaching hints and language details I may have missed. This is the first section that I felt was poorly explained. He should not have been explaining bitwise operators in terms of their integer representations, in general, such as changing a 4 into a 251. That isn’t how they are used.

Back in the dinosaur days of computers, memory was precious, and most real programming was done in assembly language. Many home computers only had 16K of RAM, and multiplication wasn’t built into the processor. If you wanted to do multiplication quickly (the computers weren’t that fast), you often wanted a precomputed multiplication table. But if you needed it for 1-100, that was 10000 entries, most of the 16K. Here’s what we did: x*y = ((x+y)^2-(x-y)^2))/4. Addition and subtraction were built into the processor, and now we only need a table of squares up to 200, so 200 entries, not 10000! And division by 4? Super easy, just do a right shift of 2, so >>2. Just like in decimal dividing by 100 corresponds to moving the decimal 2 places, dividing by 4 corresponds to moving the “decimal point” 2 places in binary, and dropping the remainder. So left and right shifts correspond to multiplying and dividing by powers of two. With optimized compilers, this isn’t as useful as it used to be.

Bitwise & and | are still used often for masks, especially graphically. For example, suppose we want a flashlight effect on the screen, and only the 4 middle bits of 8 bits should be lit. The mask 0011 1100 will turn off the bits on each side, and leave the middle 4 alone, with bitwise and:

1101 1101

&0011 1100

———-

0001 1100

And notice what the same mask does with bitwise or:

1101 1101

|0011 1100

———-

1111 1101

It turns the 4 middle bits on, and leaves the outer 4 alone.

What if we wanted to turn the high bit of a byte on, and leave the remaining bits alone, well then we do bitwise or with 1000 0000:

0101 1100

|1000 0000

———-

1101 1100

This is one of the main uses of the bitwise operators. Think how much they were used with the original black and white Mac Classic graphically. Bitwise and would dim sections of the screen, and then bitwise or could be used to draw something else in place. And bitwise or could also be used to draw a pattern on top of an existing picture in paint programs.

Good question. I added a section on using bit flags, as well as an RGBA example of using a 32-bit integer to hold four 8-bit sub-values.

Is there a logical XOR operator.

Hey joe,

If you look back on section 3.6, the 8th comment from the top of the comments is one posted by Florian in response to the 7th comment, made by RonnieTheBear.

The post begins as stated below:

Comment by Florian

2009-12-04 13:36:39

Hey Ronnie,

This should answer your question. There is no logical XOR in C++, but Florian gives an alternate option one could use to achieve the same ends.

Hi, excellent tutorial.

I’m an electronics student and I use C to program embedded systems. Now I’m looking to switch from C to C++.

In embedded systems it is really useful to have the bitwise operators. For example, bitwise operator target cpu registers and variables that control several peripheral of the cpu without altering other bits.

Also it is usefull when output pins are shared for several function in the same IO bank.

Perhaps in computers they are disappearing, but embedded systems still use them!

Save the |= and &= !!

I had been studying for hours a day, and was fine until I got to this point. For some reason, my brain isn’t grasping it. I remember running into the same problem with pointers years ago, but I finally got what they are. I must be stupid.

~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