Multiple methods to implement the value of the swap 2 int variable

The most common method is to save the backup value with a temporary variable

void swap(intint &y){    int temp = x;    x = y;    y = temp;}

Do not use temporary variables, as follows: Bitwise XOR and arithmetic implementation

#include <iostream>#include <limits>using namespace STD;voidSwapint&x,int&y) {x ^= y;    y = x ^ y; x = x ^ y;}voidSWAP1 (int&x,int&y) {x = x + y;//Even if overflow, the result is still correcty = x-y; x = XY;}voidSWAP2 (int&x,int&y) {x = XY;//x-y data lossy = x + y; x = x + y;}voidSWAP3 (int&x,int&y) {x = x * y;//x*y may overflowy = x/y; × = x/y;}voidSWAP4 (int&x,int&y) {x = y/x; y = y/x;//x = 0 Errorx = x * y;}intMainvoid){intx = numeric_limits<int>::max ();inty = numeric_limits<int>::max ()-1;//x = 1, y = 2;    cout<<"Original value"<< x <<" "<< y << Endl;    Swap (x, y); Swap (x, y);cout<<"Swap xor:";cout<< x <<" "<< y << Endl;    Swap1 (x, y); Swap1 (x, y);cout<<"swap plus:";cout<< x <<" "<< y << Endl;    SWAP2 (x, y); SWAP2 (x, y);cout<<"swap minus:";cout<< x <<" "<< y << Endl;//SWAP3 (x, y);    //cout << "swap by:";    //cout << x << "" << y << Endl;    //SWAP4 (x, y);    //cout << "swap except:";    //cout << x << "" << y << Endl;    cout<<"==========="<< Endl; x = numeric_limits<int>::min (); y = numeric_limits<int>::min () +1;cout<<"Original value"<< x <<" "<< y << Endl;    Swap (x, y); Swap (x, y);cout<<"Swap xor:";cout<< x <<" "<< y << Endl;    Swap1 (x, y); Swap1 (x, y);cout<<"swap plus:";cout<< x <<" "<< y << Endl;    SWAP2 (x, y); SWAP2 (x, y);cout<<"swap minus:";cout<< x <<" "<< y << Endl;}

The result of the operation is:

原值2147483647 2147483646以下交换使用一种方法均进行交换2次,如仍输出原值，结果正确swap 异或：2147483647 2147483646swap 加：2147483647 2147483646swap 减：-2147483647 -2147483648===========原值-2147483648 -2147483647swap 异或：-2147483648 -2147483647swap 加：-2147483648 -2147483647swap 减：2147483646 2147483647[Finished in 0.4s]

From the above results analysis, the Exchange can be implemented correctly only by bitwise XOR and by means of a variable and (although overflow but the result is correct) .

// 异或void swap(intint &y){    x ^= y;    y = x ^ y;    x = x ^ y;}// 用 x 保存 x+y 的和void swap1(intint &y){    x = x + y;    y = x - y;    x = x - y;}

Other ways, such as by saving 2 variables of the difference/product/quotient , for some data may output the correct results, but for potentially overflow data, can not achieve the correct Exchange .

Conclusion：

In addition to using temporary variables to implement the interchange method, you can also Exchange 2 integer variables in the form of bitwise XOR and with one of the variables saved and .

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Multiple methods to implement the value of the swap 2 int variable