Intermediate C++ Game Programming Tutorial 23
From Chilipedia
Bit twiddling is a lot of bullshit that is usually avoided when possible because it leads to ass code. Still, it is necessary in various situations, so you gotta learn this shit, and now is as good a time as any I guess.
Topics Covered
- Bitwise AND (&), OR (|), and shift left/right (>> and <<)
- Unary bitwise NOT/complement (~) and bitwise XOR (^)
- Masking with & and combining with |
- Packing and unpacking smaller data in larger types
- Bit flags
- Bitwise arithmetic optimizations
Video Timestamp Index
- Converting integers to their binary representations 0:25
- Introducing
std::string ToBin(unsigned int n, int min_digits=0);
- Introducing
struct EqVec2 { template <typename T> bool operator()( const T& lhs,const T& rhs ) const { return (lhs.x == rhs.x) && (lhs.y == rhs.y); } };
- Bitwise operators AND (
&
), OR (|
) 1:50
- Biswise AND (
&
): 1010 & 1100 = 1000, also called a mask - Biswise OR (
|
): 1010 | 1100 = 1110, also called a mash
- Biswise AND (
- Shifting operations left (
<<
), right (>>
) 5:10
- Shift left (
<<
): 0011 << 1 = 0110, shifts binary zeros in from the right side into the lower order bit position - Shift right (
>>
): 0110 >> 2 = 0001, shifts binary zeros in from the left side into the highter order bit position - Except for negative integers:
(-1)>>n
: ...1111>>2 = ...1111, shifts binary ones in from the left
- Shift left (
- Packing and unpacking data, masking with
&
and mashing/combining with|
7:07
- Example: combining four 8-bit numbers into one 32-bit number by using
<<
and|
- Example: unpacking a 32-bit number to an 8 bit sequence by using
&
and>>
- Example: combining four 8-bit numbers into one 32-bit number by using
- Example use case: The
Color
class in the Framework 12:20
- XRGB 8-bit values are packed into a 32-bit word using
(x<<24u) | (r<<16u) | (g<<8u) | b)
- XRGB 8-bit values are packed into a 32-bit word using
- Packing flag values into a single integer parameter 14:29
- Used in libraries such as the Windows API, DirectX, in STL
- Each value is represented by a single bit set in their binary representation
- For example:
enum Options { option1 = 0b01, option2 = 0b10 };
- Check for options with Bitwise AND: if you have an
int options
, check the option flag usingif( options & option1 )
- Combine options with Bitwise OR: make an
int options = option1 | option3;
- Advanced Bitwise operators NOT (
~
), XOR (^
) 16:48
- Bitwise NOT (
~a
): unary operation that flips all the bits ofa
- - Use case: switch off an option:
options &= ~option1;
- Bitwise XOR (
^
): binary operation sets each bit position to 1 only if the bits at that position of the two operands are different
- - Use case: a "configurable flipper".
a^0b11110000
will flip the 4 left bits ofa
and leave the 4 right bits the same - - Toggle an option:
options ^= option1;
- Bitwise NOT (
- Don't use Bitwise AND in logical operations like
if ( a & b )
19:49
-
bool(x)
evaluates to false if x==0, otherwise to true -
bool( 0b01 && 0b10 )
is true, whereasbool( 0b01 & 0b10 )
is false - Only reason why you would want to use
&
in logical operations is to avoid short circuiting evaluations with&&
or||
- - Short circuiting stops evaluating a comparison when the first argument is evaluated
- - For instance:
if ( false && ... )
will not evaluate anything after&&
because it is irrelevant - - To be safe when you do this, use
if ( bool(a) & bool(b) )
-
- Bitwise arithmetic optimizations 22:30
-
x >> n
divides x by 2^n (as in: 2 to the power of n) -
x << n
multiplies x by 2^n (as in: 2 to the power of n) -
x % 16
is equal tox & 15
orx & 0xF
- Note: no need to apply this in your code, the compiler will do this anyway when it optimizes
-
Homework
- Figure out how the binary string formatter routine works
- Figure out why
x % 16
compiles tox & 0xF
The solution for the problems will not have its own video. Feel free to come on the discord and compare your answers with others, or to ask specific questions if you are having trouble understanding.