std::list::remove 方法是否调用每个被删除元素的析构函数？

Question

我有代码：

std::list<Node *> lst;
//....
Node * node = /* get from somewhere pointer on my node */;
lst.remove(node);

std::list::remove

方法是否调用每个删除元素的析构函数（和释放内存）？如果是这样，我该如何避免？

Answer 1

是的，从容器中取出

Foo*

会破坏

Foo*

，但不会释放

Foo

。销毁原始指针“始终”是无操作的。不可能有其他方式！让我给你几个理由。 存储类

只有当指针对象实际上是动态分配的时候，删除指针才有意义，但是运行时怎么可能知道指针变量被销毁时是否是这种情况呢？指针还可以指向静态和自动变量，删除其中之一会产生

未定义的行为

。

{
    Foo x;
    Foo* p = &x;

    Foo* q = new Foo;

    // Has *q been allocated dynamically?
    // (The answer is YES, but the runtime doesn't know that.)

    // Has *p been allocated dynamically?
    // (The answer is NO, but the runtime doesn't know that.)
}

悬空指针

无法确定被指向者过去是否已经被释放。两次删除同一个指针会产生

未定义的行为

。 （第一次删除后，它变成了悬空指针。）

{
    Foo* p = new Foo;

    Foo* q = p;

    // Has *q already been released?
    // (The answer is NO, but the runtime doesn't know that.)

    // (...suppose that pointees WOULD be automatically released...)

    // Has *p already been released?
    // (The answer WOULD now be YES, but the runtime doesn't know that.)
}

未初始化的指针

根本无法检测指针变量是否已经初始化。猜猜当你尝试删除这样的指针时会发生什么？再次，答案是

未定义的行为

。

    {
        Foo* p;

        // Has p been properly initialized?
        // (The answer is NO, but the runtime doesn't know that.)
    }

动态数组

类型系统不区分指向单个对象的指针（

Foo*

）和指向对象数组的第一个元素的指针（也

Foo*

）。当指针变量被销毁时，运行时不可能确定是通过

delete

还是通过

delete[]

释放指针。通过错误的形式释放会调用

未定义的行为

。

{
    Foo* p = new Foo;

    Foo* q = new Foo[100];

    // What should I do, delete q or delete[] q?
    // (The answer is delete[] q, but the runtime doesn't know that.)

    // What should I do, delete p or delete[] p?
    // (The answer is delete p, but the runtime doesn't know that.)
}

总结

由于运行时无法对指针对象执行任何明智的操作，因此销毁指针变量是

始终

无操作。什么都不做肯定比由于不知情的猜测而导致未定义的行为要好:-) 建议

考虑使用智能指针作为容器的值类型，而不是原始指针，因为它们负责在不再需要指针时释放指针。根据您的需要，使用

std::shared_ptr<Foo>

或

std::unique_ptr<Foo>

。如果您的编译器尚不支持 C++0x，请使用

boost::shared_ptr<Foo>

。

从不

，我再说一遍，从不使用std::auto_ptr<Foo>作为容器的值类型。

Answer 2

list

中每个项目的析构函数——但这不是

Node

对象。 这是一个

Node*

。所以它不会删除

Node

指针。

这有道理吗？

Answer 3

std::list<T>::remove

将调用

T

的析构函数（当

T

类似于

std::vector

时，这是必需的）。在你的例子中，它会调用

Node*

的析构函数，这是一个空操作。它不会调用

node

的析构函数。

Answer 4

Node 上是否调用了析构函数？ 不，但“Node”不是列表中的元素类型。

至于你的另一个问题，你不能。 标准列表容器（实际上所有标准容器）采用其内容的所有权并将其清理。 如果您不希望这种情况发生，标准容器不是一个好的选择。

Answer 5

std::list

中，因此不会在指向的

Node

对象上调用析构函数。如果您想将堆分配的对象存储在 STL 容器中并在删除时销毁它们，请将它们包装在智能指针中，如

boost::shared_ptr

Answer 6

简而言之，对于类型

Node*

，既不调用解构函数，也不调用删除/释放；然而，对于

Node

类型，将调用解构函数，同时考虑删除/释放是列表的实现细节。意思是，这取决于列表实现是否使用 new/malloc。在

unique_ptr<Node>

的情况下，会调用解构函数，并且会调用删除/释放，因为您必须给它由

new

分配的东西。

#include <iostream>
#include <list>
#include <memory>

using namespace std;

void* operator new(size_t size) {
    cout << "new operator with size " << size << endl;
    return malloc(size);
}

void operator delete(void *ptr) {
    cout << "delete operator for " << ptr << endl;
    free(ptr);
}

class Apple {
public:
    int id;

    Apple() : id(0) { cout << "apple " << this << ":" << this->id << " constructed" << endl; } 
    Apple(int id) : id(id) { cout << "apple " << this << ":" << this->id << " constructed" << endl; }
    ~Apple() { cout << "apple " << this << ":" << this->id << " deconstructed" << endl; }

    bool operator==(const Apple &right) {
        return this->id == right.id;
    }

    static void* operator new(size_t size) {
        cout << "new was called for Apple" << endl;
        return malloc(size);
    }

    static void operator delete(void *ptr) {
        cout << "delete was called for Apple" << endl;
        free(ptr);
    }
    /*
        The compiler generates one of these and simply assignments
        member variable. Think memcpy. It can be disabled by uncommenting
        the below requiring the usage of std::move or one can be implemented.
    */
    //Apple& operator=(const Apple &from) = delete;
};

int main() {
    list<Apple*> a = list<Apple*>();

    /* deconstructor not called */
    /* memory not released using delete */
    cout << "test 1" << endl;
    a.push_back(new Apple());
    a.pop_back();

    /* deconstructor not called */
    /* memory not released using delete */
    cout << "test 2" << endl;
    Apple *b = new Apple();
    a.push_back(b);
    a.remove(b);
    cout << "list size is now " << a.size() << endl;

    list<Apple> c = list<Apple>();      
    cout << "test 3" << endl;
    c.push_back(Apple(1)); /* deconstructed after copy by value (memcpy like) */
    c.push_back(Apple(2)); /* deconstructed after copy by value (memcpy like) */

    /*
       the list implementation will call new... but not
       call constructor when Apple(2) is pushed; however,
       delete will be called; since it was copied by value
       in the last push_back call

       double deconstructor on object with same data
    */
    c.pop_back();

    Apple z(10);

    /* will remove nothing */
    c.remove(z);

    cout << "test 4" << endl;

    /* Apple(5) will never deconstruct. It was literally overwritten by Apple(1). */
    /* Think memcpy... but not exactly. */
    z = Apple(1);

    /* will remove by matching using the operator== of Apple or default operator== */
    c.remove(z);

    cout << "test 5" << endl;
    list<unique_ptr<Apple>> d = list<unique_ptr<Apple>>();
    d.push_back(unique_ptr<Apple>(new Apple()));
    d.pop_back();

    /* z deconstructs */
    return 0;
}

特别注意内存地址。您可以通过范围判断哪些指向堆栈，哪些指向堆。