The XOR ( ^ ) is an logical operator that will return 1 when the bits are different and 0 elsewhere. A negative number is stored in binary as two's complement. In 2's complement, The leftmost bit position is reserved for the sign of the value (positive or negative) and doesn't contribute towards the value of number.
The compiler will just produce assembly code to XOR a register onto itself). Now, if X XOR X is 0, and XOR is associative, and you need to find out what number hasn't repeated in a sequence of numbers where all other numbers have been repeated two (or any other odd number of times). If we had the repeating numbers together, they will XOR to 0.
XOR evaluation, as you understand, cannot be short-circuited since the result always depends on both operands. So 1 is out of question. But what about 2? If you don't care about 2, then with normalized (i.e. bool) values operator != does the job of XOR in terms of the result. And the operands can be easily normalized with unary !, if necessary.
XOR behaves like Austin explained, as an exclusive OR, either A or B but not both and neither yields false. There are 16 possible logical operators for two inputs since the truth table consists of 4 combinations there are 16 possible ways to arrange two boolean parameters and the corresponding output.
I am having some trouble identifying when to use the XOR operator when doing bitwise manipulations. Bitwise And and Or are pretty straight forward. When you want to mask bits, use a bitwise AND (co...
The xor operator on two booleans is logical xor (unlike on ints, where it's bitwise). Which makes sense, since bool is just a subclass of int, but is implemented to only have the values 0 and 1. And logical xor is equivalent to bitwise xor when the domain is restricted to 0 and 1. So the logical_xor function would be implemented like:
2 one is exclusive or -> xor, and the other is or which means at least one of the conditions are met in a truth table. Or is not mutually exclusive, xor is.
But a "logical" xor operator (^^) would always have to evaluate both operands. This makes it different to the other "logical" operators which evaluate the second operand only if necessary. I think this is why there is no "logical" xor in Javascript, to avoid confusion. So what should happen if both operands are falsy? Both could be returned.
It is the Bitwise xor operator in java which results 1 for different value of bit (ie 1 ^ 0 = 1) and 0 for same value of bit (ie 0 ^ 0 = 0) when a number is written in binary form. ex :- To use your example: The binary representation of 5 is 0101. The binary representation of 4 is 0100.
