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Doubt in Xor Operator

 
Ranch Hand
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Result is -
true // clear
false // clear
false // creating doubt

(As book says that ^ (xor) returns true if exactly one operand is true.) Here in the given code (at line 3) b1 is true, so it must prints (b2 ^b1) = true...... please correct me
 
Greenhorn
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The second statement (!b1 & (b2 = true)) sets b2 to true.
So b1 ^ b2 will be false since both are true.
 
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That isn't tricky at all. Did you notice in 2nd statement b2 is assigned to true? ( b2 = true , instead of b2==true ).
So in 3rd statement both b1 and b2 are true , hence answer is false !!
 
sunil langeh
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sorry guys ,,silly mistake , but thanks for quick replies
 
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