c – 可以将shared_ptr向上转换为shared_ptr导致未定义的行为吗？

共享指针非常聪明.他们记住了最初构造的类型,以便正确删除它们.以此为例：

struct A { virtual void test() = 0; };
struct B : A { void test() override {} };

void someFunc() {
    std::shared_ptr<A> ptr1;

    ptr1 = std::make_shared<B>();

    // Here at the end of the scope, B is deleted correctly
}

然而,void指针似乎存在一个问题：为了使void指针的向下转换有效,必须将它向下转换为它最初上传的类型.

例如：

void* myB = new B;

// Okay, well defined
doStuff(static_cast<B*>(myB));

// uh oh, not good!
// For the same instance of a child object, a pointer to the base and
// a pointer to the child can be differrent.
doStuff(static_cast<A*>(myB));

使用std :: shared_ptr时,当你使用std :: make_shared时,删除器看起来必须类似于这个函数：[](B * ptr){delete ptr; }.由于指针(在第一个示例中)在指向A的指针中保存B实例并正确删除它,因此必须以某种方式向下转发它.

我的问题是：以下代码片段是否调用未定义的行为？

void someFunc() {
    {
        std::shared_ptr<void> ptr = std::make_shared<B>();

        // Deleting the pointer seems okay to me,
        // the shared_ptr knows that a B was originally allocated with a B and
        // will send the void pointer to the deleter that's delete a B.
    }

    std::shared_ptr<void> vptr;

    {
        std::shared_ptr<A> ptr = std::make_shared<B>();

        // ptr is pointing to the base, which can be 
        // different address than the pointer to the child.

        // assigning the pointer to the base to the void pointer.
        // according to my current knowledge of void pointers, 
        // any future use of the pointer must cast it to a A* or end up in UB.
        vptr = ptr;
    }

    // is the pointer deleted correctly or it tries to
    // cast the void pointer to a B pointer without downcasting to a A* first?
    // Or is it well defined and std::shared_ptr uses some dark magic unknown to me?
}

解决方法:

代码是正确的.

std :: shared_ptr在内部保存真实指针和真实删除器,因为它们在构造函数中,所以无论你如何向下转换它,只要向下转换有效,删除器就是正确的.

shared_ptr实际上不包含指向对象的指针,而是指向保存实际对象,引用计数器和删除器的中间结构的指针.如果转换shared_ptr并不重要,那个中间结构不会改变.它不能改变,因为你的vptr和ptr虽然有不同的类型,但它们共享引用计数器(当然还有对象和删除器).

顺便说一下,中间结构是make_shared优化的原因：它在同一个内存块中分配中间结构和对象本身,并避免额外的分配.

为了说明智能指针是如何形成的,我编写了一个带有普通指针的程序,因为你的问题崩溃了(使用GCC 6.2.1)：

#include <memory>
#include <iostream>

struct A
{
    int a;
    A() :a(1) {}
    ~A()
    {
        std::cout << "~A " << a << std::endl;
    }
};

struct X
{
    int x;
    X() :x(3) {}
    ~X()
    {
        std::cout << "~X " << x << std::endl;
    }
};

struct B : X, A
{
    int b;
    B() : b(2) {}
    ~B()
    {
        std::cout << "~B " << b << std::endl;
    }
};

int main()
{
    A* a = new B;
    void * v = a;
    delete (B*)v; //crash!

    return 0;    
}

实际上它会输出错误的整数值,这证明了UB.

~B 0
~A 2
~X 1
*** Error in `./a.out': free(): invalid pointer: 0x0000000001629c24 ***

但智能指针的版本工作得很好：

int main()
{
    std::shared_ptr<void> vptr;

    {
        std::shared_ptr<A> ptr = std::make_shared<B>();
        vptr = ptr;
    }
    return 0;
}

它按预期打印：

~B 2
~A 1
~X 3

标签：c,shared-ptr,pointers,void-pointers
来源： https://codeday.me/bug/20190722/1502310.html