[英]converting base class pointer to unknown derived class pointer
我有一個抽象基類的5個派生類。 每個派生類中都存在一個重載的函數 ,我們將其命名為print()
。 派生4類的示例:
Derived4::print(*Derived1)
Derived4::print(*Derived2)
Derived4::print(*Derived3)
Derived4::print(*Base)
就像我之前說過的,所有派生類都具有打印功能,但參數卻不同,例如
Derived1::print(*Derived2)
Derived1::print(*Derived3)
Derived1::print(*Derived4)
Derived1::print(*Base)
所有對象都存儲在像
vector<Base*> a
當我從vector中提取其中之一並嘗試調用print函數時,所有調用都定向到print(* Base)函數。我不允許存儲類型,因此不知道vector來自什么。 ,也不允許類型檢查。
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;
}
結果:
greetings from A
greetings from A
預期結果:
greetings from C
greetings from A
您需要虛擬功能。 A::print
聲明為虛擬將使之成為事實,以便在類型為A
的指針上調用print會調用構造對象的類的print
,而不是使用指針類型來決定要調用的print
。
您還需要刪除D::print
,如果對象為D
類型,則會調用A::print
。
#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.
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