將基類指針轉換爲未知的派生類指針

Question

我從抽象基類有5個派生類。 一個函數被重載，它存在於每個派生類中的，我們將其命名爲print()。爲派生4類例如：

Derived4::print(*Derived1) 
Derived4::print(*Derived2) 
Derived4::print(*Derived3) 
Derived4::print(*Base) 

就像我以前說過，所有派生類具有打印功能，但是參數是不同的，像

Derived1::print(*Derived2) 
Derived1::print(*Derived3) 
Derived1::print(*Derived4) 
Derived1::print(*Base) 

所有對象被存儲在矢量內像

vector<Base*> a 

當我從矢量中取出其中一個並嘗試調用打印函數時，所有的調用都被指向打印（* Base）函數。我不允許存儲這些類型，因此沒有任何想法來自vector.Also，類型檢查也是不允許的。

一個例子：

#include <iostream> 
#include <vector> 
using namespace std; 
class A{ 
public: 
    void print(){cout << "greetings from A" << endl;} 
}; 

class C : public A{ 
public: 
    void print(){cout << "greetings from C" << endl;} 
}; 

class D : public A{ 
public: 
    void print(){cout << "greetings from D" << endl;} 
}; 

class B : public A{ 
public: 
    void print(C* c){c->print();} 
    void print(A* d){d->print();} 
}; 

int main() 
{ 
    D d; 
    C c; 
    B b; 
    vector<A*> a; //B,C,D will be stored inside a vector like this. 
    a.push_back(&c); 
    a.push_back(&d); 
    b.print(a[0]); 
    b.print(a[1]); 
    return 0; 
} 

其結果是：

greetings from A 
greetings from A 

期望的結果：

greetings from C 
greetings from A 

Answer 1

需要虛函數。聲明A::print爲虛函數將使得在A類型的指針上調用打印將調用構造該對象的類的print，而不是使用指針類型來決定調用哪個print。

您還需要刪除D::print像您期望A::print被調用該對象的類型是D

#include <iostream> 
#include <vector> 
using namespace std; 
class A{ 
public: 
    virtual void print(){ cout << "This is printed twice." << endl; } 
}; 

class C : public A{ 
public: 
    void print(){ cout << "This is desired output." << endl; } 
}; 

class D : public A{ 

}; 

class B : public A{ 
public: 
    void print(C* c){ c->print(); } 
    void print(A* d){ d->print(); } 
}; 

int main() 
{ 
    D d; 
    C c; 
    B b; 
    vector<A*> a; //B,C,D will be stored inside a vector like this. 
    a.push_back(&c); 
    a.push_back(&d); 
    b.print(a[0]); 
    b.print(a[1]); 
    return 0; 
} 

結果：

This is desired output. 
This is printed twice.