表达式必须修改左值答案

【问题标题】：expression must be modified lvalue表达式必须修改左值
【发布时间】：2015-03-04 04:17:18
【问题描述】：

class Base

{

  private:

    int _b;

  public:

    Base();

    Base(int b);

    virtual void display();

    //Assignment operator overload.

    Base& operator=(const Base&);

};

Base::Base()

{

    _b = 0;

}

Base::Base(int b)

{

    _b = b;

}

void Base::display()

{

    cout<<"base value := "<<_b<<endl;

}

Base& Base::operator=(const Base& ob)

{

    //Check for self-assignment.

    if(this != &ob)

    {

        this->_b = ob._b;

    }

    return *this;

}

class Derived : public Base

{

  private:

    int _d;

  public:

    Derived();

    Derived(int d);

    void display();

    //Assignment operator overload.

    Derived & operator=(const Derived& ob);

};

Derived::Derived() : Base()

{

    _d = 0;

}

Derived::Derived(int d) : Base(d)

{

    _d = d;

}

void Derived::display()

{

    cout<<"Derived value := "<<_d<<endl;

}

Derived & Derived::operator=(const Derived& ob)

{

    if(this != &ob)

    {

        this->_d = ob._d;

    }

    return *this;

}

int main()

{

    Derived d1(10),d2(),d3;

    //How d2 becomes lvalue and not d3 above.

    d2 = d1;//Error :: expression must be modified lvalue.

    //d2.display();

    d3 = d1;

    return 0;

}

【问题讨论】：

你有什么问题？
问问题的时候看预览了吗？如您所见，它非常不可读。请修正格式。
d2() 声明了一个函数（不带参数并返回 Derived），请注意。
谢谢，我想我会使用默认构造函数调用。

标签： c++ operator-overloading lvalue

【解决方案1】：

Derived d2();

被视为函数声明。这样做：-

Derived d1(10),d2,d3;
d2 = d1;          /////ahaa it's working

【讨论】：

【解决方案2】：

问题是你的'd2()'。语句 d2() 不会像您想象的那样创建一个名为 d2 的派生对象。正确的选项是：

Derived d2;
Derived d2 = Derived();
Derived d2(0); // or any other integer

声明

Derived d2();

告诉编译器您正在定义一个不带参数并返回 Derived 类型的函数。

更多解释请参见此处： error: request for member '..' in '..' which is of non-class type

【讨论】：