c++ - 使用返回类型作为派生类的指针

Question

是否有可能在基类中有一个函数，它将返回类型作为指向派生类的指针？主要目的是供以后在使用 Base 和 Derived 类设置值时使用。

Base B1;
B1.SetName("nameOfBase");
Derived* D1 = B1.CreateDerived("DerivedFromBase");//CreateDerived method will be in class Base
D1->SetMinPoint(0,1);//this method will be implemented in derived class
D1->SetMaxPoint(4,4);//this method will be implemented in derived class

我在实现中遇到了问题，我做了类似的事情

class Base
{
public:
  Base();
  bool SetName(char*);//or SetName(string)
  Derived* CreateDerived(char*); // or Derived* CreateDerived(string)
  ~Base();
protected:
  char baseName[20];// or string baseName
  Derived* derivedPtr[5];
};

class Derived: public Base
{
public:
  Derived();
  bool SetName(char*);//the name given in Derived* CreateDerived(char*) will be set here
  ~Derived(); 
};

当我尝试执行此操作并运行程序时，出现类似

的错误

// Derived* CreateDerived(char*); // or Derived* CreateDerived(string)
error C2143: syntax error: missing ';' before '*'
error C4430: missing type identifier: int assumed.

Answer 1

是的，这是可能的。这似乎是一个有点可疑的设计，但没有什么能阻止你这样做。您只需要在引用它之前声明类 Derived:

class Derived;  //forward declaration here

class Base
{
public:
  Base();
  bool SetName(char*);//or SetName(string)
  Derived* CreateDerived(char*); // or Derived* CreateDerived(string)
  ~Base();
protected:
  char baseName[20];// or string baseName
  Derived* derivedPtr[5];
};

class Derived: public Base
{
public:
  Derived();
  bool SetName(char*); //don't forget semi-colon
  ~Derived(); 
};

正如@psur 提到的，您还缺少 SetName 后的分号。

而且我强烈建议在 C++ 中对字符串使用 std::string 而不是 char*。