Bitwise operations and shifts

Question

Im having some trouble understanding how and why this code works the way it does. My partner in this assignment finished this part and I cant get ahold of him to find out ho

Answer 1

c = 33 + ~n;

This calculates how many high order bits are remaining after using n low order bits.

((x << c)>>c

This fills the high order bits with the same value as the sign bit of x.

!(blah ^ x)

This is equivalent to

blah == x