`void operator=(T&&)` 和 `T& operator=(T&&)` 有什么区别？

Question

class base 
{
public:
    base(const base&) = delete;
    base()
    {
        cout << "construct" << endl;
    }
    ~base() 
    {
        cout << "destruct" << endl;
    }

    int a;
    int b;

    /* The difference explanation I desired is here */
    void operator=(base&& other)  
    // base& operator=(base&& other) // this needs to collaborate with "return *this" 
    {
        this->a = other.a;
        this->b = other.b;
        // return *this;
    }

    /* Not here */
    base& operator=(base& other) = delete;
};

operator=(T&&)两个版本有什么区别？他们似乎都对我有用。但是，作为class成员函数，本站推荐base& operator=(T&&)版本。

Answer 1

在一种情况下，a=b=c 有效。另一方面，它没有。

就是这样。

传统上，a=b=c 执行 b=c，然后将结果分配给 a。如果你的 operator= returns void，它反而编译失败。

`void operator=(T&&)` 和 `T& operator=(T&&)` 有什么区别？

What is the difference between `void operator=(T&&)` and `T& operator=(T&&)`?

c++

operator-overloading

return-value

assignment-operator