混乱关于C++类参考

Question

下面的代码是正确的

string s1="abc"; 
    string s2="bcd"; 
    string &rs1=s1; 
    string &rs2=s2; 
    rs1=rs2; 
    cout<<rs1<<"----"<<rs2<<endl; 

与以下几部件代码将编译错误：

class A 
{ 
public: 
    A(string& a):ma(a) { } 

    string& ma; 
}; 

string s1="abc"; 
string s2="bcd"; 
A oa(s1); 
A ob(s2); 
oa=ob; 
cout<<oa.ma<<"----"<<ob.ma<<endl; 

以上数据string&类型分配，为什么把它们放到类会引起编译错误？ （gcc版本4.7.1）

错误信息是

non-static reference member 'std::string& A::ma', can't use default assignment operator 

Answer 1

你设定一个成员引用本地临时变量。在你的构造函数中的参数是一个temp）。这导致了“悬挂参考”，这是不好的。

将param更改为引用，或将您的成员更改为非引用。为了您的目的，你可能会想：

class A 
{ 
public: 
    A(string& a):ma(a) { } 

    A& operator =(const A& other) 
    { 
     ma = other.ma; 
     return *this; 
    } 

    string& ma; 
}; 

但是你应该知道，你的类的默认复制构造函数可能不会做你认为它会。

UPDATE

处理为什么上课的时候有一个参考成员的默认拷贝赋值运算符将被删除标准的具体领域：

C++11 § 12.8,p23

A defaulted copy/move assignment operator for class X is defined as deleted if X has: - a variant member with a non-trivial corresponding assignment operator and X is a union-like class, or

• a non-static data member of const non-class type (or array thereof), or
• a non-static data member of reference type, or
• a non-static data member of class type M (or array thereof) that cannot be copied/moved because overload resolution (13.3), as applied to M’s corresponding assignment operator, results in an ambiguity or a function that is deleted or inaccessible from the defaulted assignment operator, or
• a direct or virtual base class B that cannot be copied/moved because overload resolution (13.3), as applied to B’s corresponding assignment operator, results in an ambiguity or a function that is deleted or inaccessible from the defaulted assignment operator, or
• for the move assignment operator, a non-static data member or direct base class with a type that does not have a move assignment operator and is not trivially copyable, or any direct or indirect virtual base class.