[Solved] Simple logic operation >> &

Question

That code isolates the two bits indicated by the arrows below, and moves them to the two least significant bit positions.

            vv
11111111111111111111111111111111 original value
                              vv
00000000000000000011111111111111 after >>18 (shift right 18 positions)
                              vv
00000000000000000000000000000011 after & 3 (mask out all but the 0th and 1st bits)

This assumes an unsigned value, and no special handling of the sign bit during the shift. The & (AND) is a masking operation because for the result to be 1 at any position, both the value and the mask must be 1 (i.e. 1 AND 1 = 1, all else is 0. So only the bits in the mask (the 0x03 in your case) can end up with anything other than 0, and they will only be 1 if they were 1 in the value being masked.

Accepted Answer

That code isolates the two bits indicated by the arrows below, and moves them to the two least significant bit positions.

            vv
11111111111111111111111111111111 original value
                              vv
00000000000000000011111111111111 after >>18 (shift right 18 positions)
                              vv
00000000000000000000000000000011 after & 3 (mask out all but the 0th and 1st bits)

This assumes an unsigned value, and no special handling of the sign bit during the shift. The & (AND) is a masking operation because for the result to be 1 at any position, both the value and the mask must be 1 (i.e. 1 AND 1 = 1, all else is 0. So only the bits in the mask (the 0x03 in your case) can end up with anything other than 0, and they will only be 1 if they were 1 in the value being masked.