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What are the properties of XOR?
XOR is one of the sixteen possible binary operations on Boolean operands. That means that it takes 2 inputs (it’s binary) and produces one output (it’s an operation), and the inputs and outputs may only take the values of TRUE or FALSE (it’s Boolean) – see Figure 1….All About XOR.
| XOR Truth Table | ||
|---|---|---|
| Input A | Input B | Output |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
How do you swap two variables in XOR?
Method 2 (Using Bitwise XOR) The bitwise XOR operator can be used to swap two variables. The XOR of two numbers x and y returns a number that has all the bits as 1 wherever bits of x and y differ. For example, XOR of 10 (In Binary 1010) and 5 (In Binary 0101) is 1111 and XOR of 7 (0111) and 5 (0101) is (0010).
Is Bitwise XOR unique?
The way to find the unique element is by using the XOR bitwise operator which returns 1 only if one of the element is 1 and false otherwise. In the loop, each of the integers in the array are XORed with uniqueId starting at 0. Then, 0 is XOR’ed with 34.
When does the swap by Xor ALgorithm work?
When both references refer to the same variable, this method zeroes it. Wikipedia has an excellent explanation of the Swap-By-XOR algorithm. The formal proof that this algorithm works is a bit involved, and requires the use of the mathematical properties of binary numbers.
Do you need temporary storage for XOR swap?
Conventional swapping requires the use of a temporary storage variable. Using the XOR swap algorithm, however, no temporary storage is needed. The algorithm is as follows: The algorithm typically corresponds to three machine-code instructions.
Is it safe to use XOR swap in high level language?
Therefore, XOR swapping must not be used blindly in a high-level language if aliasing is possible.
How do you swap two sets of bits?
We need to swap two sets of bits. XOR can be used in a similar way as it is used to swap 2 numbers. Following is the algorithm. 1) Move all bits of the first set to the rightmost side set1 = (x >> p1) & ((1U << n) – 1) Here the expression (1U << n) – 1 gives a number that contains last n bits set and other bits as 0.