Question

我有一个练习来编写一个替换没有第三个变量的值的函数，我尝试了一些解决方案，但没有得到它们。

我写了以下代码：

void swap1(int* p1, int* p2) {
    p1++;
    *p1 = *p2;

    p1--;
    *p2 = *p1;

    p1++;    
    cout << *p1 << endl;
}

主要：

    int main() {
    int n1 = 5, n2 = 8;

    cout << "before swap: " << endl;
    cout << "value 1: " << n1 << " value 2: " << n2 << endl;

    swap1(&n1, &n2);

    cout << "after swap: " << endl;
    cout << "value 1: " << n1 << " value 2: " << n2 << endl;

    return 0;
}

我得到以下输出：

before swap:
value 1: 5 value 2: 8
8
after swap:
value 1: 5 value 2: 5
Press any key to continue . . .

为什么当我在我的函数中打印时，这实际上交换了值，但在main中没有？

怎么了？

*p1 = p1++;＆gt;＆gt;为什么不工作？（我在理论上尝试做的事）

Answer 1

没有溢出。

请参阅http://www.geeksforgeeks.org/swap-two-numbers-without-using-temporary-variable/

 int main() {

    int n1 = 5, n2 = 8;


    cout << "before swap: " << endl;
    cout << "value 1: " << n1 << " value 2: " << n2 << endl;

     n2 = n1^n2;
     n1 = n1^n2;
     n2 = n1^n2;

    cout << "after swap: " << endl;
    cout << "value 1: " << n1 << " value 2: " << n2 << endl;

    return 0;
}

Answer 2

正如之前的答案中所建议的那样......只需按照您提议的形式输入：

void swap2(int* p1, int* p2) {
  if(p1 != p2) {
    *p2 = (*p1)^(*p2);
    *p1 = (*p1)^(*p2);
    *p2 = (*p1)^(*p2);
  }
}

Answer 3

 int main() {

    int n1 = 5, n2 = 8;


    cout << "before swap: " << endl;
    cout << "value 1: " << n1 << " value 2: " << n2 << endl;

     n2 = n1+n2;
     n1 = n2 -n1;
    n2 = n2-n1;


    cout << "after swap: " << endl;
    cout << "value 1: " << n1 << " value 2: " << n2 << endl;

    return 0;
}

在没有其他变量的情况下替换值的程序

3 个答案: