c++ - 通过迭代器反转 vector 的元素

Question

我尝试编写以下函数:

typedef std::vector<int>::iterator v_itt;

 void reverse(v_itt first, v_itt second)
 {
         std::cout << "first = " << *first << ", second = " << *second << std::endl;
         int *temp = &*first;
         *second = *first;
         *first = *temp;
         std::cout << "After reversing: first = " << *first << ", second = " << *second << std::endl;
 }

但是当我尝试调用它时，我得到了以下输出:

first = 322, second = 12
After reversing: first = 322, second = 322

我想，当我们用另一个指针初始化一个指针时，在初始化的指针中我们只是得到一个值的拷贝。这不是真的吗？

Answer 1

要交换元素你应该这样做:

temp = first
first = second // Here you have "second = first". It's the other way around!
second = temp

您在第二步中犯了一个错误，这导致您的 reverse 操作为两者都提供了 first。关键思想是:您将值 (first) 存储到 temp，然后通过以下操作可以覆盖 that 值 (first)，因为你不会失去它。