静态成员函数和运行时间多态

Question

class Base
{
    private:
    Base() = default;
    static Base *b;
    public:
    static Base* get();
};
class Derived: public Base
{

};
Base* Base::b=nullptr;
Base* Base::get(){Base* b = new Derived();return b;}
void main()
{
     Base* b = Base::get();  
}

我遇到编译时错误：

main.cpp: In static member function 'static Base* Base::get()':
main.cpp:14:41: error: use of deleted function 'Derived::Derived()'
   14 | Base* Base::get(){Base* b = new Derived();return b;}
      |                                         ^
main.cpp:9:7: note: 'Derived::Derived()' is implicitly deleted because the default definition would be ill-formed:
    9 | class Derived: public Base
      |       ^~~~~~~
main.cpp:9:7: error: 'constexpr Base::Base()' is private within this context
main.cpp:4:5: note: declared private here
    4 |     Base() = default;
      |     ^~~~

Live Example

在Base::get函数中，如果我做Base* b = new Base();或删除私有 Base() 构造函数并使其成为 public，我没有收到任何错误。

Answer 1

通过将 Base() 构造函数设为私有，Derived() 默认构造函数变得格式错误（它试图调用私有 Base()），因此被隐式删除。然后您尝试使用它来构造一个 Derived，这会给出您看到的错误。

为了使这项工作正常进行，需要有一个 Derived() 构造函数——您可以自己定义一个构造函数，也可以安排默认构造函数不存在错误格式。您可以通过使 Base() public 或 protected 而不是 private 来实现后者（因此 Derived() 构造函数可以调用它）。

所有与静态成员和虚函数无关的东西。

静态成员函数和运行时间多态

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