如果我有指向它的指针而不是迭代器，如何删除矢量元素？

Question

我遇到了 child object 需要 parent 来销毁它的问题。我想要类似下面的东西，这只是一个例子：

#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;
}

如果我跟踪 child 在矢量中的位置，我知道该怎么做，但我基本上想知道如果我有指向的指针，如何删除矢量的元素它。

编辑：AndyG 一直都是对的，我不能做我想做的事，因为当我 "new" 它时，我的 Objects 在内存中到处都是。我确实设法用另一种方式使用 placement new，在一个连续的缓冲区中创建 objects，但这绝对不值得麻烦。不过我确实学到了很多。

#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 
        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，唯一的方法就是像往常一样搜索向量。

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

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

Demo

Answer 2

假设vector不需要排序，那么我们可以将child元素交换到末尾，然后调整vector的大小。此方法不需要将向量的所有元素移动到 child 和最后一个元素之间。

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;
}

如果我有指向它的指针而不是迭代器，如何删除矢量元素？

How can I delete a vector element if I have a pointer to it, not an iterator?

c++

iterator

vector

erase

Demo