This week's book giveaway is in the Mac OS forum. We're giving away four copies of a choice of "Take Control of Upgrading to Yosemite" or "Take Control of Automating Your Mac" and have Joe Kissell on-line! See this thread for details.
int n = 7; n<<=3;<br /> n = n & n + 1 | n + 2 ^ n + 3;<br /> n >>= 2; System.out.println(n); The order of execution for the bitwise and or and Xor operator is as follows first it evaluates the bitwise and(&) then it evaluates the bitwise Xor(^) ann then it evaluates the bitwise or(|) but in the expression(n = n & n + 1 | n + 2 ^ n + 3 )) all the airthmetic operators will be evaluated first then the bitwise operators and finally the assignment took place. so the result is 56. 8 | 9 & 10 ^ 11 result will be 14. boolean b1=b2=b3=b4=true b1 | b2 & b3 ^ b4 you can not do assignment and initialization at the same time. rather you can say boolean b1,b2,b3,b4; b1=b2=b3=b4=true; b1 | b2 & b3 ^ b4; then result would be true. & is first to be evaluated then ^ and then | same resion for the following. b4 = b4 | b1 & b2; b3 = b3 & b1 | b2; correct me if i am wrong. Thanks. [This message has been edited by Amit, Jhalani (edited January 12, 2001).] [This message has been edited by Amit, Jhalani (edited January 12, 2001).]