c++ - std::list::remove 方法是否调用每个已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 方法是否调用每个已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，删除 Foo*从容器中销毁 Foo* ，但它不会释放 Foo .销毁原始指针总是是无操作的。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 undefined behavior .

{
    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 undefined behavior . (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

根本无法检测指针变量是否已初始化。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 undefined behavior .

    {
        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 (也是 Foo* )。当指针变量被销毁时，运行时可能无法判断是否通过 delete 释放指针。或通过delete[] .通过错误的形式释放调用 undefined behavior .

{
    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执行任何明智的操作，因此销毁指针变量总是是无操作的。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> .

Never ，我再说一遍，永远不要使用 std::auto_ptr<Foo>作为容器的值类型。