Byte swap question (Bitwise operation)

Question

I have seen some bitwise operation practice questions (in C language) and there was one I didn't understand: you had to swap the nth and the mth byte in a specifc integer "x" using only bitwise operations, and that n and m were both <=3. The solution involved masking the last 8 bits of the number resulting from xor based on x >> (n<<3) to x >> (m<<3), but the part I don't understand is why they shifted m/n right 3.

Answer 1

Because if n is a position within the word, in bytes, then 8*n is the same position in bits. And n<<3 is the same as 8*n (2³ = 8).

With n=1, n<<3 is 8, and (x >> (n << 3)) & 0xff would move the second lowest byte to the bottom position, and mask the rest of the word out. (The bytes would numbered starting from zero, of course.)

_{(I'm assuming bytes of 8 bits, since I think that's by far the most common case at least nowadays.)}