如果我有一個指向矢量元素而不是迭代器的指針，該如何刪除？

Question

我的子對象有一個問題，需要其父對象銷毀它。 我想要以下內容，這只是一個示例：

#include <iostream>
#include <vector>

struct Object
{
    Object(Object* parent) : parent(parent) {}
    Object* parent;
    std::vector<Object*> children;
    bool flag = false;
    void update() { if (flag) parent->deleteChild(this); } // Or mark it for deletion afterwards
    void deleteChild(Object* child) { delete child; /*children.erase(/* I need the iterator here);*/ }
};

int main()
{
    Object* parent = new Object(nullptr);
    for (int i = 0; i < 100; ++i) parent->children.push_back(new Object(parent));

    parent->children[42]->flag = true;

    for (auto i : parent->children) i->update();

    return 0;
}

如果我跟蹤孩子在向量中的位置，我會知道該怎么做，但是我基本上想知道如果我有一個指向向量的元素，該如何擦除它的元素。

編輯：AndyG一直以來都是對的，我無法做我想要的事情，因為當我“新建”對象時，我的對象就在內存中了。 我確實設法使用new new放置以另一種方式進行操作，將所有對象都創建在一個連續的緩沖區中，但這絕對不值得麻煩。 我確實學到了很多東西。

#include <iostream>
#include <vector>

struct Object
{
    Object(Object* parent, int position) : parent(parent), numberPosition(position)
    {
        std::cout << "Constructing object number: " << numberPosition << " at at heap memory location: " << this << '\n';
    }

    Object* parent;
    int numberPosition = 0;
    std::vector<Object*> children;
    bool flag = false;
    void update() 
    { 
        if (flag) parent->deleteChild(this); 
    } 
    void deleteChild(Object* child) 
    { 
        Object* pChild = &(*child);
        ptrdiff_t position = child - *children.data();
        std::vector<Object*>::iterator it = children.begin() + position;
        std::cout << "About to delete vector element at position: " << (*it)->numberPosition << '\n';

        // delete pChild;   Not supposed to deallocate each placement new. See http://www.stroustrup.com/bs_faq2.html#placement-delete and https://stackoverflow.com/questions/222557/what-uses-are-there-for-placement-new
        std::cout << "Size of children vector = " << children.size() << '\n';
        children.erase(it);
        std::cout << "Size of children vector = " << children.size() << '\n';
    }
    ~Object() { std::cout << "Destroying object number " << numberPosition << '\n'; }
};

int main()
{
    Object* parent = new Object(nullptr, 0);
    char* contiguousBuffer = static_cast<char*>(malloc(100 * sizeof(Object)));
    for (int i = 0; i < 100; ++i)
    {
        Object* newAddress = new (contiguousBuffer + i * sizeof(Object)) Object(parent, i); // Placement new
        parent->children.push_back(newAddress);
    }

    parent->children[42]->flag = true;

    //for (auto i : parent->children) i->update();  // Iterator gets invalidated after erasing the element at 42 doing it this way
    for (int i = 0; i < parent->children.size(); ++i) parent->children[i]->update();


    free(contiguousBuffer); 
    // Destructors also need to be called

    return 0;
}

Answer 1

不幸的是，唯一的方法就是像普通一樣搜索向量。

auto it = std::find(std::begin(children), std::end(children), child);

if (it != std::end(children)){
   children.erase(it);
   delete child;
}

演示版

Answer 2

假設不需要對向量進行排序，那么我們可以將child元素交換到末尾，然后調整向量的大小。 此方法不需要在child元素和最后一個元素之間移動向量的所有元素。

auto it = std::find(std::begin(children), std::end(children), child);

if (it != std::end(children)){
    std::iter_swap(children.rbegin(), it);
    children.resize(children.size() - 1);  
    delete child;
}