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.
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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.
The first method is to use hexadecimal literals.
Related content
We talk about hexadecimal in lesson 5.4 -- Literals.
Here’s how hexadecimal converts to binary:
| Hexadecimal | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | A | B | C | D | E | F |
| Binary | 0000 | 0001 | 0010 | 0011 | 0100 | 0101 | 0110 | 0111 | 1000 | 1001 | 1010 | 1011 | 1100 | 1101 | 1110 | 1111 |
Therefore, we can define bit masks using hexadecimal like this:
Sometimes leading hexadecimal 0s will be omitted (e.g. instead of 0x01 you’ll just see 0x1). Either way, this can be a little hard to read if you’re not familiar with hexadecimal to binary conversion.
An easier method is to use the left-shift operator to shift a single 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
Let’s examine how this works.
In the case of flags & mask0, we have 0000'0101 & 0000'0001. Let’s line these up:
0000'0101 & 0000'0001 --------- 0000'0001
We are then casting 0000'0001 to a bool. Since any non-zero number converts to true and this value has a non-zero digit, this evaluates to true.
In the case of flags & mask1, we have 0000'0101 & 0000'0010. Let’s line these up:
0000'0101 & 0000'0010 --------- 0000'0000
Since a zero value converts to false and this value has only zero digits, this evaluates to false.
Setting a bit
To set (turn on) a bit (to value 1), 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 reset (clear) a bit (to value 0), 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:
As an aside…
Some compilers may complain about a sign conversion with this line:
flags &= ~mask2;
Because the type of mask2 is smaller than int, operator~ causes operand mask2 to undergo integral promotion to type int. Then the compiler complains that we’re trying to use operator&= where the left operand is unsigned and the right operand is signed.
If this is the case, try the following:
We discuss integral promotion in lesson 10.2 -- Floating-point and integral promotion.
Flipping a bit
To toggle (flip) a bit state (from 0 to 1 or from 1 to 0), 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 separate booleans values would normally take 8 bytes. But the examples above (using std::uint8_t) use 9 bytes -- 8 bytes to define the bit masks, and 1 byte 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 saved using bit flags 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 argument corresponds to which option (is setting the “edit flag” the 9th, 10th, or 11th parameter?).
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:
This is much more readable.
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 |
| 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.
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.
Summary
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
Do not use std::bitset in this quiz. We’re only using std::bitset for printing.
Given the following program:
a) Add a line of code to set the article as viewed.
Expected output:
00000101
b) Add a line of code to check if the article was deleted.
c) Add a line of code to clear the article as a favorite.
Expected output (Assuming you did quiz (a)):
00000001
1d) Extra credit: why are the following two lines identical?