[英]C++ accesing pointer elements of a vector with iterator
使用下面显示的代码,我创建了4个类(3个子类和一个超类),每个类具有相同的print_info()函数。 我已经从每个类创建了指针对象,并将它们插入到使用超类创建的向量中。
#include <iostream>
#include <string>
#include <vector>
#include <iterator>
using namespace std;
class Vehicle
{
private:
int wheels, max_speed;
string model;
public:
virtual void print_info()
{
cout << "This is function print_info() in Vehicle (parent class)" << endl;
}
};
class Car: public Vehicle
{
private:
int doors, passengers;
public:
void print_info()
{
cout << "This is function print_info() in Car (child class)" << endl;
}
};
class Bike: public Vehicle
{
private:
string owner;
bool hasLights;
public:
string my_owner();
};
class Truck: public Vehicle
{
private:
float cargo_weight;
int allowed_speed;
public:
void print_info()
{
cout << "This is function print_info() in Truck (child class)" << endl;
}
};
int main()
{
Vehicle *v1 = new Vehicle();
Vehicle v2;
Car c1;
Car c2;
Bike b1;
Truck t1;
Truck t2;
vector <Vehicle> vvec;
Vehicle *v = new Vehicle();
Car *c = new Car();
Bike *b = new Bike();
Truck *t = new Truck();
vvec.push_back(*v);
vvec.push_back(*c);
vvec.push_back(*b);
vvec.push_back(*t);
vector<Vehicle>::iterator iter = vvec.begin();
while( iter != vvec.end())
{
iter->print_info();
iter++;
}
v1 = &v2;
v1->print_info();
v1 = &c2;
v1->print_info();
v1 = &t2;
v1->print_info();
system("pause");
return 0;
}
所需的输出是这样的:
这是Vehicle(父类)中的函数print_info
这是Car(子类)中的函数print_info
这是Vehicle(父类)中的函数print_info
这是Truck(子类)中的函数print_info
这是Vehicle(父类)中的函数print_info
这是Car(子类)中的函数print_info
这是Truck(子类)中的函数print_info
相反,我得到:
这是Vehicle(父类)中的函数print_info
这是Vehicle(父类)中的函数print_info
这是Vehicle(父类)中的函数print_info
这是Vehicle(父类)中的函数print_info
这是Vehicle(父类)中的函数print_info
这是Car(子类)中的函数print_info
这是Truck(子类)中的函数print_info
我相信这是由于早期绑定所致,但我不知道该如何解决!
您需要创建指针vector <Vehicle*> vvec;
,当您像vvec.push_back(*v);
一样push_back时vvec.push_back(*v);
您通过值传递元素并将其转换为Vehicle
,因此不会有多态性,因为vector的所有元素只是简单的Vehicle对象,请更改为:
//...
vector <Vehicle*> vvec;
Vehicle *v = new Vehicle();
Car *c = new Car();
Bike *b = new Bike();
Truck *t = new Truck();
vvec.push_back(v);
vvec.push_back(c);
vvec.push_back(b);
vvec.push_back(t);
vector<Vehicle*>::iterator iter = vvec.begin();
while( iter != vvec.end())
{
(*iter)->print_info();
iter++;
}
//...
您的输出将是:
This is function print_info() in Vehicle (parent class)
This is function print_info() in Car (child class)
This is function print_info() in Vehicle (parent class)
This is function print_info() in Truck (child class)
This is function print_info() in Vehicle (parent class)
This is function print_info() in Car (child class)
This is function print_info() in Truck (child class)
这是对象切片的经典示例。
当取消引用指针以将其插入向量vvec.push_back(*v)
,会将其转换为基类类型。 因此,它现在表现为“车辆”。
如果要获得预期的行为,请尝试使用vector <Vehicle*> vvec
和vvec.push_back(v)
。
您想要的是多态性,因此您必须存储Vehicle引用或Vehicle指针。 而且,由于您不能将引用存储在vector
,因此要使用指针。
vector <Vehicle*> vvec;
然后只需按下指针并调用print_info
将触发多态行为
vvec.push_back(v);
vvec.push_back(c);
vvec.push_back(b);
vvec.push_back(t);
vector<Vehicle*>::iterator iter = vvec.begin();
while( iter != vvec.end())
{
(*iter)->print_info();
iter++;
}
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