================================================== Exercise 2.7-Setting bits at a position n Inverted ================================================== Question ======== Write a function invert(x,p,n) that returns x with the n bits that begin at position p inverted (i.e., 1 changed into 0 and vice versa), leaving the others unchanged. .. literalinclude:: cprogs/ex_2.7_invert.c :language: c Explanation =========== Let's take n as 3:: ~(~0 << n) = ~(~0 << 3) = ~(1111 1111 << 3) = ~(1 1111 000 << 3) = ~(1111 1000) = 0000 0111 This sets the last n bits to 1 and rest to 0 The given position p 0 indexed and the number of bits, n is 1 indexed. So in order to change the bits starting at position, p, we need to shift our streams 1s to position p and we accomplish that by left shifting the stream of 1s to position (p+1-n) For the position 3 We get:: ~(~0 << n) << (p+1 -n) = 0000 0111 << (3+1 -3) = 0000 0111 << 1 = 000 0111 0 = 0000 1111 Inverting the bits can accomplished by the XOR operation ``(^)`` So, a value like 8 which is 0000 1000 can be inverted can inverted from 3rd position onwards by:: x ^ (~(~0 << n) << (p + 1 - n)) = 0000 1000 ^ 0000 1111 = 0000 1111 = 15 Visualization ============= .. raw:: html