擦除元素在while循环中迭代

Question

我尝试在使while循环迭代的过程中删除元素。但是似乎在擦除一个元素之后，循环会跳过下一个元素。代码如下

 Example program
#include <iostream>
#include <string>
#include <deque>
#include <algorithm>
using namespace std;

deque<int*> queue;
int main()
{
  int * a1 = new int(0);
  int * a2 = new int(2);
  int * a3 = new int(3);
  int * a4 = new int(4);
  int * a5 = new int(5);
  queue.push_back(a1);
  queue.push_back(a2);
  queue.push_back(a3);
  queue.push_back(a4);
  queue.push_back(a5);

    deque<int * >::iterator it = queue.begin();

    while(it != queue.end())
    {
       std::cout << "Element :" << **it << std::endl;
       //remove all elements are divided by 2
       if((**it&1) == 0)
       {
        deque<int * >::iterator cur = (it+1);
        std::cout << "Erase " << **it << std::endl;
        queue.erase(std::remove(queue.begin(), queue.end(), *it), queue.end());
        delete *it;
        it = cur;
       }
       else
       {
           ++it;
       }
    }
    std::cout << "End" << std::endl;     
}

输出如下

*元素：0清除0

元素：3

元素：4清除4

结束*

似乎元素2和5被跳过。

Answer 1

您具有未定义的行为，因为您要取消引用delete *it;处的无效迭代器。要解决您眼前的问题，您需要类似

int * to_del = *it;
it = queue.erase(it);
delete to_del;

但是，您还应该避免在int *足够的情况下使用int，并且在<algorithm>中有适用内容时写循环。>

bool is_even(int i) { return (i % 2) == 0; }

std::deque<int> queue { 0, 2, 3, 4, 5 };
queue.erase(std::remove_if(queue.begin(), queue.end(), is_even), queue.end());