C++ meaning |= and &=

Question

I have a part of code that contains the following functions:

void Keyboard(int key)
{
    switch (key) {
    case GLFW_KEY_A: m_controlState |= TDC_LEFT; bre         


        

Answer 1

x |= y is equivalent to x = x|y

In general, for any binary operator *, a *= b is equivalent to a = a*b

If you want to know what & and | are, read about bitwise operators.

Answer 2

&= |= operators in a sense are similar to +=/-= (i.e. a &= b is equivalent to a = a & b). But, they do binary operations. & is doing bitwise and operation, while | is doing bitwise or operation.

Example:

a = 1101

b = 1011

a & b = 1001

a | b = 1111

Answer 3

x |= y specifically turns on (sets to 1) those bits in x present in y, leaving all the others in x as they were.

x &= ~y specifically turns off (sets to 0) those bits in x present in y, leaving all the others in x as they were.

Answer 4

These two:

m_controlState |= TDC_LEFT
m_controlState &= ~TDC_LEFT

are equivalent to:

m_controlState = m_controlState | TDC_LEFT
m_controlState = m_controlState & ~TDC_LEFT

It works like this with all builtin X= operators.

m_controlState is most likely treated as a bitset. m_controlState may be, e.g. 01010000 (realistically, it will be larger than 8 bits).

1) | is bitwise or, which is equivalent to addition to that bitset.

So if TDC_LEFT is 00000010:

01010000 | 00000010 = 01010010

2) ~ is bitwise negation:

~00000010 = 111111101

And if you do 01010010 & ~(00000010) = 01010000, it's effectively equivalent to bitset difference.

In short:

bitsetA + bitsetB <=> bitsetA | bitset
bitsetA - bitsetB <=> bitsetA & ~ bitset

Answer 5

it's a simple way to set & reset specific bit in a word.

bitword |= MASK is a shorthand of bitword = bitword | MASK which sets the mask bit

bitword &= ~MASK clears that bit (check your boolean algebra notebook to see why)

Answer 6

These are bitwise AND and OR operations. The lines you mentioned are equivalent to:

m_controlState = m_controlState | TDC_LEFT;
m_controlState = m_controlState & ~TDC_LEFT