C ++继承成员函数使用静态变量

时间:2012-12-23 12:34:17

标签: c++ inheritance static-members

我正在尝试将一些Python类转换为c ++但是遇到了一些麻烦。我有一个Base类,它有一个类(静态)变量和一个返回它的方法。我还有一个派生类,它会覆盖类(静态)变量,如此,

在Python中:

class Base:
   class_var = "Base"
   @classmethod
   def printClassVar(cls):
      print cls.class_var

class Derived(Base):
   class_var = "Derived"

d = Derived()
d.printClassVar()

打印出所需的派生类变量“Derived”。知道如何在c ++中获得相同的功能吗?我试过但最终得到了Base类的类变量。

在c ++中

class Base
{
public:
    static void printStaticVar(){cout << s_var << endl;}
    static string s_var;
};
string Base::s_var = "Base";

class Derived : public Base
{
public:
    static string s_var;
};
string Derived::s_var = "Derived";

void main()
{
    Derived d;
    d.printStaticVar();
}

2 个答案:

答案 0 :(得分:3)

编写一个虚函数,返回对静态成员的引用:

class Base
{
public:
    void printStaticVar() {cout << get_string() << endl;}
    static string s_var;
    virtual string const& get_string() { return Base::s_var; }
};
string Base::s_var = "Base";

class Derived : public Base
{
public:
    static string s_var;
    virtual string const& get_string() { return Derived::s_var; }
};
string Derived::s_var = "Derived";

void main()
{
    Derived d;
    d.printStaticVar();
}

请注意printStaticVar不应该是静态的。


您还可以在getter中将字符串static设为本地:

class Base
{
public:
    void printStaticVar() {cout << get_string() << endl;}
    virtual string const& get_string() { 
        static string str = "Base";
        return str;
    }
};

class Derived : public Base
{
public:
    virtual string const& get_string() { 
        static string str = "Derived";
        return str;
    }
};

void main()
{
    Derived d;
    d.printStaticVar();
}

答案 1 :(得分:0)

另一种可能性可能是:

class Base
{
  const std::string var;
public:
  Base(std::string s="Base") : var(s) {}
  void printVar() { std::cout << var << std::endl }
};
class Derived : public Base
{
public:
  Derived(std::string s="Derived") : Base(s) {}
};