向量迭代器不可取消（尝试手动反向向量）

Question

I'm trying to create a function which takes in a vector and simply reverses (manually). I'm aware of the existence of reverse(), but I ran into the "Vector iterator not dereferencable" problem and for educational purposes, I'd like to know what it means. I tried researching this problem and someone (on this forum) said that vect.end() is not dereferencable by definition, but from my understanding, using reverse_iterator is just reversing the ends, so following the logic; vect.rend should not be dereferencable.

vector<int> reverseVector(vector<int>);

int main()
{
    vector<int> vec;

    for (int i = 0; i < 11; i++)
    {
        vec.push_back(i);
    }

    vec = reverseVector(vec);

    for (vector<int>::iterator it = vec.begin(); it != vec.end(); it++)
    {
        cout << *it << " ";
    }
    cout << endl;

    return 0;
}

vector<int> reverseVector(vector<int> vect) 
{
    vector<int>::reverse_iterator ritr;
    for (ritr = vect.rbegin(); ritr != vect.rend(); ritr++)
    {
        vect.insert(vect.begin(), *ritr);
        vect.pop_back();
    }
    return vect;
}

Answer 1

Your problem has nothing to do with the dereferencability or otherwise of rend() . You're modifying the vector while iterating over it, which invalidates the iterator.

To answer your original question, a reverse_iterator isn't just "reversing the ends" compared to a forward iterator. rbegin() is end() - 1 , and rend() is begin() - 1 .

Answer 2

If you add an element to the vector, ritr may be invalidated thus the error

Vector iterator not dereferencable.

Thus its better using an index as your loop variable or better a copy(temp) vector for reverse task.

Answer 3

You are deleting elements from the vector (popping from the back), which invalidates the reverse iterator.

---

You could just iterate through half of the vector and swap the elements, lke this:

void swap(int& a, int& b) {
    int tmp = a;
    a = b;
    b = tmp;
}

vector<int> reverseVector(vector<int> vect) {
    const size_t origin_size = vect.size();
    for(size_t i = 0; i < origin_size/2; ++i)
        swap(vect[i], vect[origin_size - 1 - i]);
    return vect;
}

Answer 4

insert和pop_back成员函数都修改了向量并使迭代器无效。

Answer 5

A tip as design-issue: use always const-reference in a function, unless you really know what you are doing. So you will avoid stepping in traps like this. For example:

vector<int> reverseVector(const vector<int> &vect)

Now you wont have this problem, because you can not modify vect.