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    Define a macro for logical XOR operation

    - Anonymous on October 06, 2010 Report Duplicate | Flag
    Microsoft Software Engineer / Developer Algorithm

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previous comment was bitwise xor
for logical xor would be
srry i didnt read the question correctly
#define XOR(a,b) (((a)&&(!(b)))||(((!a))&&(b)))

- geeks on July 24, 2011 | Flag Reply
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Or simply, a != b

- Anonymous on February 28, 2013 | Flag
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a||b?(a&&b?false:true):false

- Anonymous on October 06, 2010 | Flag Reply
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(a&^b) | (^a&b)

- Anonymous on October 06, 2010 | Flag Reply
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(!!a) ^ (!!b)

- ajay on October 09, 2010 | Flag Reply
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It's equivalent to a ^ b isn't it? That doesn't really answer the question...

- Anonymous on October 14, 2010 | Flag
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(a&&~b)||(~a&&b)

- Just on October 09, 2010 | Flag Reply
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(a&&!b)||(!a&&b)

- Anonymous on October 10, 2010 | Flag Reply
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Can anybody highlight why we cannot use simple a^b ??

- Ankur on October 11, 2010 | Flag Reply
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lol..

- Erik on October 11, 2010 | Flag
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Logical xor, not bitwise...

- Anonymous on February 28, 2013 | Flag
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The Question was we must not use any if else logic and obvious ur not supposed to use ^ operator

- Anonymous on October 15, 2010 | Flag Reply
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a XOR b = ( a | b ) & ( !a | !b )

- erm on October 23, 2010 | Flag Reply
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a XOR b = (a & !b) | (!a & b)

- b on November 16, 2010 | Flag Reply
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# define ex_or(a,b) (a^b)

- Anonymous on November 22, 2010 | Flag Reply
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Hilarious.. :)

- lego.mk on December 01, 2010 | Flag
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#define XOR(a,b) (((a)&(!(b)))|((!(a))&b))

- geeks on July 24, 2011 | Flag Reply
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previous comment is not correct
#define XOR(a,b) (((a)&(~(b)))|((~(a))&b))

- geeks on July 24, 2011 | Flag Reply


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