C++智能指针类

Question

#include <iostream>

template <typename T>
class my_smart_ptr {
    T* raw;
    unsigned * p_count;
public:
    my_smart_ptr(T* t):
    raw(t),
    p_count (new unsigned(1))
    {

    }

    my_smart_ptr(const my_smart_ptr<T> & p):
    raw(p.raw),
    p_count(p.p_count)
    {
        ++(*p_count);
    }

    const my_smart_ptr<T> & operator=(const my_smart_ptr<T>& p)
    {
        if (p == *this)
        {
            return *this;
        }

        raw = p.raw;
        --(*p_count);
        ++(*(p.p_count));
        return *this;

    }

    ~my_smart_ptr()
    {
        --(*p_count);
        if (*(p_count) == 0)
        {
            delete raw;
        }
    }
};


class A{
public:

    A()
    {
        std::cout<< "A()" << std::endl;
    }

    ~A()
    {
        std::cout<< "~A()" << std::endl;
    }

};
int main()
{
    my_smart_ptr<A> a (new A);
    return 0;
}

Answer 1

operator=中有一个错误。当函数返回时，LHS的p_count成员仍然指向旧的p_count。它应该等于RHS的p_count。此外，当计数达到零时，忘记释放旧数据。

const my_smart_ptr<T> & operator=(const my_smart_ptr<T>& p)
{
    if (p == *this)
    {
        return *this;
    }

    // Keep a copy of the old values.
    // Don't change these until the state of the
    // object has been made consistent.
    unsigned*  oldCount = p_count;
    T*         oldRaw   = raw;

    // Update the state of the object.
    // Making sure to get the local count correct.
    raw = p.raw;
    p_count = p.p_count;  // Without this the use count is incorrect.
    ++(*(p.p_count));

    // Now clean up the old state.
    --*oldCount;
    if (*oldCount == 0) {
        delete oldCount;
        delete oldRaw;
    }
    return *this;
}

operator=函数通常返回一个非const引用。毕竟，您只是修改了对象。

my_smart_ptr<T> & operator=(const my_smart_ptr<T>& p) { ... }