即使应用了重置，智能指针仍然引用原始指针

Question

请找到我从https://www.geeksforgeeks.org/auto_ptr-unique_ptr-shared_ptr-weak_ptr-2/获取的用于测试智能指针的代码。

// C++ program to demonstrate shared_ptr
#include <iostream>
#include <memory>

class A {
public:
    void show()
    {
        std::cout << "A::show()" << std::endl;
    }
};

int main()
{
    std::shared_ptr<A> p1(new A);
    std::cout << p1.get() << std::endl;
    p1->show();
    std::shared_ptr<A> p2(p1);
    p2->show();
    std::cout << p1.get() << std::endl;
    std::cout << p2.get() << std::endl;

    // Returns the number of shared_ptr objects
    // referring to the same managed object.
    std::cout << p1.use_count() << std::endl;
    std::cout << p2.use_count() << std::endl;

    // Relinquishes ownership of p1 on the object
    // and pointer becomes NULL
    p1.reset();
    std::cout << p1.get() << std::endl;
    std::cout << p2.use_count() << std::endl;
    std::cout << p2.get() << std::endl;
    p1->show();
    p2->show();
    std::cout << p1.get() << std::endl;
    std::cout << p2.use_count() << std::endl;
    std::cout << p2.get() << std::endl;

    return 0;
}

p1 被重置并且没有指针分配给它（根据 p1.get() 找到）。之后，当我调用 p1->show() 函数时，它将输出显示为 A::show()。怎么可能呢？原始指针也是同样的情况吗？

output:
0x24dc5ef1790
A::show()
A::show()
0x24dc5ef1790
0x24dc5ef1790
2
2
0
1
0x24dc5ef1790
A::show() 
A::show()
0
1
0x24dc5ef1790

Answer 1

它之所以有效，是因为您没有使用对象中的任何成员，否则会出现段错误。但消毒剂很容易发现这个问题：

$ g++ -ggdb -O0 -fsanitize=undefined,address shared.cpp -o shared
$ ./shared
0x602000000010
A::show()
A::show()
0x602000000010
0x602000000010
2
2
0
1
0x602000000010
shared.cpp:33:13: runtime error: member call on null pointer of type 'struct element_type'
A::show()
A::show()
0
1
0x602000000010

更改代码以添加一个成员变量以进行打印，如下所示

class A {
public:
    void show()
    {
        std::cout << "A::show() " << value << std::endl;
    }
    int value;
};

它会出现段错误

$ g++ -ggdb -O0 shared.cpp -o shared
$ ./shared
0x560a3dde4eb0
A::show() 0
A::show() 0
0x560a3dde4eb0
0x560a3dde4eb0
2
2
0
1
0x560a3dde4eb0
Segmentation fault (core dumped)