c++ - ‘operator=’没有匹配项

在下面的示例中，我尝试如果尝试从fun()引用返回，那么它会警告我。

但是，当我尝试按值返回时，我期望'operator=()'被触发，但是它抛出了以下错误，我无法找出原因:

error: no match for ‘operator=’ (operand types are ‘Auto_ptr2<Resource>’ and ‘Auto_ptr2<Resource>’)

note: no known conversion for argument 1 from ‘Auto_ptr2<Resource>’ to
‘Auto_ptr2<Resource>&’

令人惊讶的是，res2=res1调用了'operator =()'就好了!!!

template<class T>
class Auto_ptr2
{
        T* m_ptr;
public:
        Auto_ptr2(T* ptr=nullptr)
                :m_ptr(ptr)
        {std::cout<<"\nAuto_ptr CTOR called ";}
        ~Auto_ptr2()
        {
                std::cout<<"\n~Auto_ptr2 called\n";
                delete m_ptr;
        }
        // A copy constructor that implements move semantics
        Auto_ptr2(Auto_ptr2&
        {
                std::cout<<"\nAuto_PTR2 copy constructor called";
                m_ptr = a.m_ptr;
                a.m_ptr = nullptr;
        }
        Auto_ptr2& operator=(Auto_ptr2& a)
        {
                std::cout<<"\nAuto_ptr2 operator = called";
                if (&a == this)
                        return *this;
                delete m_ptr;
                m_ptr = a.m_ptr;
                a.m_ptr = nullptr;
                return *this;
        }

        T& operator*() const { return *m_ptr; }
        T* operator->() const { return m_ptr; }
        bool isNull() const { return m_ptr == nullptr;  }
};
class Resource
{
public:
        Resource() { std::cout << "\nResource acquired"; }
        ~Resource() { std::cout << "\nResource destroyed"; }
};

Auto_ptr2<Resource> fun(Auto_ptr2<Resource> res3)
{
        std::cout << "\nres1 is " << (res3.isNull() ? "null" : "not null");
        return res3;
}

int main()
{
        Auto_ptr2<Resource> res1(new Resource);
        Auto_ptr2<Resource> res2;

        res2 = res1; // res2 assumes ownership, res1 is set to null

        res2=fun(res2); //ERROR : error: no match for ‘operator=’ (operand types are ‘Auto_ptr2<Resource>’ and ‘Auto_ptr2<Resource>’)


        return 0;
}

最佳答案

如果我理解正确，那么您正在编写类似于std::unique_ptr的内容。

因此，对于std::unique_ptr，operator=()应该可以使用move语义，因此接收r值引用Auto_ptr &&而不是l值引用

     Auto_ptr2& operator=(Auto_ptr2 && a)
    {
            std::cout<<"\nAuto_ptr2 operator = called";
            if (&a == this)
                    return *this;
            delete m_ptr;
            m_ptr = a.m_ptr;
            a.m_ptr = nullptr;
            return *this;
    }

并且使用应通过std::move()

res2=fun(std::move(res2));

构造函数存在相同的问题:避免复制构造函数(可能将其删除)并编写move构造函数。

    Auto_ptr2 (Auto_ptr2 const &) = delete;

    Auto_ptr2 (Auto_ptr2 && a)
    {
            std::cout<<"\nAuto_PTR2 move constructor called";
            m_ptr = a.m_ptr;
            a.m_ptr = nullptr;
    }