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问题:

解答:

#include <iostream>
using namespace std;void swap1(int& a, int& b)
{int temp = a;a = b;b = temp;
}void swap2(int& a, int& b)
{a = a + b;b = a - b;a = a - b;
}void swap3(int& a, int& b)
{a = a ^ b;b = a ^ b;a = a ^ b;
}int main()
{int a1 = 1, b1 = 2;int a2 = 3, b2 = 4;int a3 = 5, b3 = 6;int a = 2147483647, b = 1;swap1(a1, b1);swap2(a2, b2);swap3(a3, b3);printf("after swap. . .\n");printf("a1 = %d,b1 = %d\n", a1, b1);printf("a2 = %d,b2 = %d\n", a2,b2); printf("a3 = %d,b3 = %d\n", a3, b3); swap2(a,b);printf("a = %d,b = %d\n", a, b);return 0;
}

运行结果:

注意:

• 方法二可能会溢出,所以方式三是最佳解.

记下来吧…