27

我有代码：

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

该std::list::remove方法是否调用每个已删除元素的析构函数（和空闲内存）？如果是这样，我该如何避免它？

4

6 回答 6

47

是的，Foo*从容器中删除 a 会破坏Foo*，但不会释放Foo. 销毁原始指针始终是无操作的。不能有别的办法！让我给你几个理由。

存储类

删除指针只有在指针对象实际上是动态分配的情况下才有意义，但是运行时如何知道指针变量被销毁时是否是这种情况？指针也可以指向静态和自动变量，删除其中一个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

没有办法确定指针是否在过去已经被释放。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一个悬空指针。）

{
    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

也根本无法检测指针变量是否已被初始化。猜猜当您尝试删除这样的指针时会发生什么？再一次，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Foo*的指针 () 和指向对象数组的第一个元素的指针 (also Foo*)。当指针变量被销毁时，运行时可能无法确定是通过delete还是通过释放指针delete[]。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.)
}

概括

由于运行时无法对指针做任何有意义的事情，因此销毁指针变量始终是无操作的。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>。如果您的编译器还不支持 C++0x，请使用boost::shared_ptr<Foo>.

从不，我再说一遍，永远不要用作std::auto_ptr<Foo>容器的值类型。

于 2010-11-23T21:20:41.550 回答
12

它调用中每个项目的析构函数list-- 但这不是一个Node对象。它的一个Node*。

所以它不会删除Node指针。

那有意义吗？

于 2010-11-23T20:19:10.350 回答
7

它确实调用了列表中数据的析构函数。这意味着，std::list<T>::remove将调用的析构函数（在类似的情况下T是必要的）。Tstd::vector

在您的情况下，它将调用 的析构函数Node*，这是一个无操作。它不调用node.

于 2010-11-23T20:20:41.080 回答
3

是的，尽管在这种情况下，Node* 没有析构函数。但是，根据其内部结构，不同的 Node* 值会被范围规则删除或销毁。如果 Node* 中有一些非基本类型，则将调用析构函数。

析构函数是否在节点上调用？不，但“节点”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（实际上是所有标准容器）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，那么标准容器不是一个好的选择。

于 2010-11-23T20:20:58.123 回答
0

由于您将指针放入 a std::list，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，请将它们包装在一个智能指针中，例如boost::shared_ptr

于 2010-11-23T20:18:53.083 回答
0

最好的理解方法是测试每个表格并观察结果。要巧妙地将容器对象与您自己的自定义对象一起使用，您需要对行为有很好的理解。

简而言之，对于Node*既不调用解构函数也不调用 delete/free 的类型；但是，对于该类型Node，将调用解构器，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了 new/malloc。

在 a 的情况下unique_ptr<Node>，将调用解构器并且调用 delete/free 将发生，因为您必须给它分配由new.

#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;
}

请特别注意内存地址。您可以通过范围判断哪些指向堆栈，哪些指向堆。

于 2016-08-17T18:15:39.870 回答