如果我有一个指向矢量元素的指针,而不是迭代器,如何删除它?

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

本文关键字:何删除 迭代器 删除 有一个 指针 元素 如果      更新时间:2023-10-16

我遇到了一个子对象需要其父对象来销毁它的问题。我想要如下所示的内容,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;
}

如果我跟踪孩子在矢量中的位置,我就知道如何去做,但我基本上想知道如果我有一个指向矢量的指针,我如何擦除矢量的元素。

编辑:AndyG一直都是对的,我不能做我想做的事,因为当我"新"它时,我的对象在内存中到处都是。我确实设法使用放置 new 以另一种方式做到这一点,在一个连续的缓冲区中创建所有对象,但这绝对不值得麻烦。不过,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;
}

不幸的是,唯一的方法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;
}

演示

假设向量不需要排序,那么我们可以将child元素交换到末尾,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;
}