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; break;
    case GLFW_KEY_D: m_controlState |= TDC_RIGHT; break;
    case GLFW_KEY_W: m_controlState |= TDC_UP; break;
    case GLFW_KEY_S: m_controlState |= TDC_DOWN; break;
    default: Test::Keyboard(key);
    }
}

void KeyboardUp( int key)
{
    switch (key) {
    case GLFW_KEY_A: m_controlState &= ~TDC_LEFT; break;
    case GLFW_KEY_D: m_controlState &= ~TDC_RIGHT; break;
    case GLFW_KEY_W: m_controlState &= ~TDC_UP; break;
    case GLFW_KEY_S: m_controlState &= ~TDC_DOWN; break;
    default: Test::Keyboard(key);
    }
}

I know what a switch case is but i dont understand what these parts do.

m_controlState |= TDC_LEFT
m_controlState &= ~TDC_LEFT

m_controlState is an int. The GFLW_KEY's also refere to an int value.

Could someone explain what these parts do an example with input values and results would be nice.

Answer 1

Also I think it should be explained what these operators do and are used this way.

m_controlState serves as flags, which means it contains in binary form which of the keys are pressed. For example if the values of tds constants are chosed like this:

TDS_LEFT             = 0x00001
TDS_RIGH = 0x01 << 2 = 0x00010 
TDS_UP   = 0x01 << 3 = 0x00100
TDS_DOWN = 0x01 << 4 = 0x01000

Then in single integer you can store information which options are set. To do that you just have to check if bit that corresponds on each setting is 1 or 0.

So to set TDS_LEFT option, you have to OR the current state with 0x00001( which is TDS_LEFT), so in code

m_controlState = m_controlState | TDS_LEFT

which is the same as

m_controlState |= TDS_LEFT.

To unset TDS_LEFT option you have to AND it with ~TDS_LEFT. So

m_controlState = m_controlState & ~TDS_LEFT

which is the same as:

m_controlState &= ~TDS_LEFT

You can also check: How to use enums as flags in C++?. Hope that makes it clearer.

Answer 2

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 3

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

Answer 4

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 5

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 6

&= |= 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