Question: Please explained the codes N /* * bitxor - x^y using only and & Example: bitxor (4, 5) = 1 Legal ops: ~ & Max

Please explained the codes N /* * bitxor - x^y using only Please explained the codes

N /* * bitxor - x^y using only and & Example: bitxor (4, 5) = 1 Legal ops: ~ & Max ops: 14 Rating: 1 */ int bitxor(int x, int y) { return ~(x & y) & ~(~x &~ y); }

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