Question

Example

byte b = 127;

(initialize be to equal 11111111)

Now I only care about bits 1 and 0

how do i bit shift do get 00000011

I think this is called “masking bits” if I’m right?

I tried
b << 5
then
b>> 5
to zero out the other bits but that’s just wrong

My goal

switch ((myByte >> 3) & 3)
{
    case 3:
        resevered = true;
        break;
    case 2:
        open = true;
        break;
    case 1:
        SingleOnly = true;
        break;
    case 0:
        daulMode = true;
        break;
}

Answer 1

b & 0x03

will give you a byte containing the first two bits, with the remaining bits zero-filled.

This works because 0x03 is 00000011 in binary, and

11111111

AND

00000011

zeroes the first six bits, leaving only the remaining two bits.