std :: list :: remove方法是否调用每个被删除元素的析构函数？

Question

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

std :: list :: remove方法是否调用每个被删除元素的析构函数(和空闲内存)？如果是,我怎么能避免呢？

Answer 1

是的,Foo*从容器中删除a 会破坏Foo*,但不会释放Foo.销毁原始指针总是无操作.它不能是任何其他方式!让我举几个理由.

存储类

如果指针实际上是动态分配的,那么删除指针才有意义,但运行时如何知道指针变量被销毁时是否就是这种情况？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.)
}

晃来晃去的指针

无法确定指针对象是否已在过去发布.删除相同的指针两次会产生未定义的行为.(它在第一次删除后变成悬空指针.)

{
    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

也无法检测指针变量是否已经初始化.猜猜当你试图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(也Foo*).当指针变量被销毁时,运行时无法确定是否通过delete或通过释放指针delete[].通过错误的表单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.)
}

摘要

由于运行时无法对指针对象做任何事情,因此销毁指针变量始终是无操作.无所事事肯定比由于不知情的猜测导致未定义的行为更好:-)

忠告

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时释放指针.根据您的需要,使用std::shared_ptr<Foo>或std::unique_ptr<Foo>.如果你的编译器还不支持C++ 0x,请使用boost::shared_ptr<Foo>.

从来没有,我再说一遍,永远不要使用std::auto_ptr<Foo>作为容器的值类型.

Answer 2

它调用了每个项目的析构函数list- 但这不是一个Node对象.它是一个Node*.

所以它不会删除Node指针.

那有意义吗？

Answer 3

它确实调用列表中数据的析构函数.这意味着,std::list<T>::remove将调用析构函数T时(这是必要的T是一样的东西std::vector).

在你的情况下,它会调用析构函数Node*,这是一个无操作.它不会调用析构函数node.