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

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