只是试图绕过我如何释放作为参数传递的新对象的内存。
假设我有一个像这样定义的Link类:
class Link{
public:
Link(const std::string& value, Link* previous = nullptr, Link* successor = nullptr) : _value{ value }, _previous{ previous }, _successor{ successor }{}
Link* Insert(Link* new_link);
Link* getPreviousLink() const{ return _previous; }
Link* getSuccessorLink() const{ return _successor; }
std::string _value;
private:
Link* _previous;
Link* _successor;
};
然后我就像这样定义Link * Insert(Link *):
Link* Link::Insert(Link* new_link){
if(!new_link){ return this; }
if(!this){ return new_link; }
new_link->_successor = this;
if(_previous){ _previous->_successor = new_link; }
new_link->_previous = _previous;
_previous = new_link;
return new_link;
}
然后在我的main()中,我执行以下操作:
int main(){
Link* Cars = new Link("Ford");
Cars = Cars->Insert(new Link("Ferrari"));
Cars = Cars->Insert(new Link("Hummer"));
Cars = Cars->Insert(new Link("Volvo"));
}
我创建了一个名为“Cars”的链接指针,并在堆上分配了一个值为“Ford”的新链接。然后我将我的Cars指针指向从Insert()返回的新链接。再重复此步骤2次。
我的问题是,当我将新的Link对象作为参数传递时,如何删除或释放分配的内存?我是否在Link的析构函数中这样做?如果我只删除我的汽车指针,它就不会释放其他链接。
答案 0 :(得分:4)
使用智能指针,所有权将很明确:
class Link : public std::enable_shared_from_this<Link> {
public:
Link(const std::string& value) : _value{ value } {}
std::shared_ptr<Link> Insert(std::shared_ptr<Link> new_link);
std::shared_ptr<Link> getPreviousLink() const{ return _previous.lock(); }
std::shared_ptr<Link> getSuccessorLink() const{ return _successor; }
std::string _value;
private:
std::weak_ptr<Link> _previous;
std::shared_ptr<Link> _successor;
};
std::shared_ptr<Link> Link::Insert(std::shared_ptr<Link> new_link)
{
if (!new_link){ return shared_from_this(); }
new_link->_successor = shared_from_this();
auto prev = _previous.lock();
if (prev) { prev->_successor = new_link; }
new_link->_previous = prev;
_previous = new_link;
return new_link;
}
int main(){
auto Cars = std::make_shared<Link>("Ford");
Cars = Cars->Insert(std::make_shared<Link>("Ferrari"));
Cars = Cars->Insert(std::make_shared<Link>("Hummer"));
Cars = Cars->Insert(std::make_shared<Link>("Volvo"));
}