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8= 00000000 00000000 00000000 00001000
~8 = 11111111 11111111 11111111 11110111
we are assigning it in integer hence most significant bit (MSB)is the sign bit
bcz MSB is 1 hence it is treated as -ve no.
when u try to print it then before printing compiler will take 2's complement hence it becomes :
2's complement of (~8)=9
2's complement of 11111111 11111111 11111111 11110111 is 00000000 00000000 00000000 00001000 +1 = 1001 = 9

~ operator is, by definition, the operator to negate all bits, so it's the most appropriate answer.
FFFF is 16 bits, Xor with 0xFFFFFFFF would work

Question is to negate each bit ,i.e 1000 should become 0111

the operator is nothing but eg:-) if input is 8 then (8+1)*(-1) would get the ~ result.. correct me if its wrong:)

the answer must b -9

first of all, its output will be -9. (not 9)
if we take int is of 1 byte, then 8= 0000 1000
its ~, ~8=1111 0111
now compiler sees, oh, its MSB is 1, so it must be negative number. So in order to print it, I will take 2's compliment of this binary combination, find its decimal equivalent, put negative sign before it, and print it, That's it !
decimal equivalent of 1111 0111= -(2's comp of 1111 0111)
= -(0000 1001)
= -(9)

what is the question?

ideone.com/u5udnP does this answer your question.Basically two's complement is used to represent negative numbers.