Bitwise Operation in C: Understanding x = (x | x << 2) & 0x33

The expression "x = (x | x << 2) & 0x33" performs bitwise operations on the variable "x". 1. "x << 2" shifts the bits of "x" two positions to the left. 2. "x | x << 2" performs a bitwise OR operation between "x" and the shifted value of "x". - This operation sets the two bits that were shifted to 1 if they were originally 1 in "x". 3. "& 0x33" performs a bitwise AND operation between the result of the previous step and the value "0x33" (which is equivalent to binary "00110011"). - This operation clears all bits in the result that are not set in "0x33". - It effectively keeps the lower two bits and the upper two bits of the result, while setting all other bits to 0. 4. Finally, the result is assigned back to "x". In summary, this expression updates "x" by setting the two bits that are two positions to the left of the original value of "x", while keeping the lower two bits and the upper two bits unchanged.