Question

class A
{
public:
    int i;
    ~A()
    {   
        std::cout << "~A" << std::endl;
    }   
};

class B: public A
{
public:
    int k;
    ~B()
    {   
        std::cout << "~B" << std::endl;
    }   
};

int main(int argc, char* argv[])
{
    A* p = new B();
    delete p;
    return 0;
}

上述情况不会导致内存泄漏,尽管基底破坏器不是虚拟的,而且我知道原因。

但是,这是没有定义的行为吗?

说即使基地破坏器是非虚拟的,如果衍生类不指向其他动态数据,就不会有内存泄漏?

Answer 1

是。通过一个指针将导出 class 的物体删除到没有 irtual 毁灭器的基类是教科书 UB。

<强度>5.3.5/3:

In the first alternative (delete object), if the static type of the operand is different from its dynamic type, the static type shall be a base class of the operand’s dynamic type and the static type shall have a virtual destructor or the behavior is undefined. In the second alternative (delete array) if the dynamic type of the object to be deleted differs from its static type, the behavior is undefined.73)

因为它是未定义的行为, 猜测代码是否会泄漏并不合理。只是试图修正代码, 而不是预测结果。

Answer 2

是,因为静态类型(此处为 A )不同于动态类型(您示例中的 B ),而且没有 irtual 毁灭器,它是一个UB。

5.3.5/2 (C+/11中的5.3.5/3):

In the first alternative (delete object), if the static type of the operand is different from its dynamic type, the static type shall be a base class of the operand’s dynamic type and the static type shall have a virtual destructor or the behavior is undefined.

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