{"id":436825,"date":"2018-07-25T04:36:05","date_gmt":"2018-07-25T04:36:05","guid":{"rendered":"https:\/\/essaypaper.org\/an-example-of-swap-without-temp\/"},"modified":"2018-10-24T09:10:37","modified_gmt":"2018-10-24T09:10:37","slug":"an-example-of-swap-without-temp","status":"publish","type":"post","link":"https:\/\/www.benedictsol.com\/blogs\/an-example-of-swap-without-temp\/","title":{"rendered":"An Example of Swap Without Temp"},"content":{"rendered":"
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Here is a solution to the problem of<\/span><\/i> swap without temp<\/i>. Now a similar problem that a student may have can be solved as follows in our example. If our sample has made a good impression on you then you should consider getting help from Assignment.EssayShark.com. Our expert will follow all of your requirements and academic standards while dealing with your order. Next, we should mention that we are available 24\/7, so you can get our help any time you need.<\/span><\/i><\/p>\n

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Swap the Values of Variables Without Use of the Temporary Variables <\/span><\/p>\n

This is a classic problem. It is usually offered in interviews, and it is quite simple. Assume x0 is the original value of x,<\/span> and y0 is the original value of y. Assume dif as a difference between x and y (x \u2013 b). Let\u2019s show the relative position of all of these values on the real axis in the case when x > y. <\/span><\/p>\n

\"\"<\/p>\n

Assign x the value of dif. If we add the value of y and dif, we will get x0 and save the result to y. Now we have y = x0 and x = dif. Now all we have to do is assign y the value of x0 \u2013 dif, which is equal to y \u2013 x.<\/span><\/p>\n

The following code implements this algorithm:<\/span><\/p>\n


\npublic static void swap(int x, int y) {
\n    \/\/for example, let x = 10, y = 6;
\n    x = x - y; \/\/ x = 10 - 4 = 6
\n    y = x + y; \/\/ y = 6 + 4 = 10
\n    x = y - x; \/\/ x = 10 - 6
\n 
\n    System.out.println(\"x: \" + x);
\n    System.out.println(\"y: \" + y);
\n}
\n<\/pre>\n
We can also solve this problem using a bit manipulation. This approach will allow us to work with different types of data, not just integers.<\/p>\n

\npublic static void swap(int x, int y) {
\n    \/\/for example, x = 101 (binary), y = 110
\n    x = x ^ y; \/\/ x = 101^110 = 011
\n    y = x ^ y; \/\/ y = 011^110 = 101
\n    x = x ^ y; \/\/ x = 011^101 = 110
\n 
\n    System.out.println(\"x: \" + x);
\n    System.out.println(\"y: \" + y);
\n}
\n<\/pre>\n
This code uses the operation XOR. The easiest way to understand the code is to look at the two bits a and b. Let\u2019s denote the initial values as a, a0, and b0.
If we will be able to swap a couple of bits, the algorithm will work correctly. Let\u2019s look at the work of the algorithm step by step:<\/p>\n
    \n
  1. a = a0 ^ b0 (0 if a0 = b0, 1 if a0 ! = b0)<\/span><\/li>\n
  2. b = a ^ b0 (equal to the value of a0)<\/span><\/li>\n
  3. a = a ^ b (equal to b0)<\/span><\/li>\n<\/ol>\n
    In line 1 we perform a = a0^b0 operation, which will be equal to 0 if a0 = b0, and equal to 1 if a0 != b0.<\/span><\/p>\n
    In line 2 we perform the b = a ^ b0 operation. Let\u2019s analyze both possible values of a. Since we want to swap the values of a and b, we want to get 0 as a result:<\/span><\/p>\n