Question

我遇到了一个需要父对象销毁子对象的问题。我想要类似下面的内容，它只是一个例子：

#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;
}

如果我跟踪孩子在向量中的位置，我知道该怎么做，但我基本上想知道如果我有一个指向它的指针我怎么能擦除向量的元素。

编辑：AndyG一直都是对的，我不能做我想要的事情，因为我的对象在记忆中到处都是我和＃34;新的＆＃34;它。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对象，但它绝对不值得麻烦。我确实学到了很多东西。

#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，唯一的方法是像正常一样搜索矢量。

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

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

Demo

Answer 2

假设不需要对向量进行排序，那么我们可以将child元素交换到结尾，然后调整向量的大小。此方法不需要在child和最后一个元素之间移动向量的所有元素。

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;
}

如果我有一个指向它的指针而不是迭代器，我怎么能删除一个向量元素？

2 个答案:

Demo