c++ - A pair of function call operators

Question

I have a function object A with a pair of function call operators (line 4 and 5):

class A{
public:
    A(int x) : _x(x){}
    int  operator () () const { return _x; }  // line 4
    int & operator () () { return _x; }   // line 5
private:
    int _x; 
};

Similar pair of call operators is used here. The question is: do I need line 4 at all? Does the operator defined at line 4 will ever by called? In the following case:

A a(7);
a() = 8;
cout << a() << endl;

always the operator from line 5 is called.

Answer 1

是的，将使用第 4 行，例如：

 A a(3);  
 const A b(2);
 a(); // from line 5
 b(); // from line 4

Answer 2

int  operator () () const { return _x; }

当你的对象是const.

还返回一个引用不是最好的设计，它打破了数据隐藏规则，set/get函数是更好的选择。当调用第 4 行或调用第 5 行时，您会感到困惑。

我建议重写为：

class A{
public:
    explict A(int x) : x_(x) {}
    //int  operator () () const { return x_; }  // leave operator() for functor.
    operator int() const { return x_; }         // use conversion function instead 
    void setX(int x) { x_ = x; }
private:
    int x_;   //suggest use trailing `_`
};