对象切片出现在多态处理对象时传值引起的。
传地址之所以不会出现切片因为:地址都相同的长度,派生类对象的地址与基类对象有相同的地址,基类可以透明的操作派生类对象
例如:
#include <stdio.h>
#include <iostream>
using namespace std;
class Pet{
string pName;
public:
Pet(const string& name) :pName(name){}
virtual string name() const{ return pName; }
virtual string describtion() const {
return "this is " + pName;
}
};
class Dog:public Pet
{
string favoriteActivity;
public:
Dog(const string& name, const string& activity) :Pet(name), favoriteActivity(activity){}
string describtion() const {
return Pet::name() + "likes to" + favoriteActivity;
}
};
void describtion(Pet p)
{
string aa = p.describtion();
cout << aa.c_str() << endl;
}
int main()
{
Pet p("Alfred");
Dog g("fluffy", "sleep");
describtion(p);
describtion(g);
getchar();
return 0;
}
运行结果:
this is Alfred
this is fluffy
此时两次均调用了基类,并未调用派生类,1.因为按值传递时编译器知道确切的类型,派生类对象被强转为基类。
2.按值传递时,基类的构造函数被调用,基类构造函数初始化vptr上的vtable,只拷贝基类部分。
当传地址时:
#include <stdio.h>
#include <iostream>
using namespace std;
class Pet{
string pName;
public:
Pet(const string& name) :pName(name){}
virtual string name() const{ return pName; }
virtual string describtion() const {
return "this is " + pName;
}
};
class Dog:public Pet
{
string favoriteActivity;
public:
Dog(const string& name, const string& activity) :Pet(name), favoriteActivity(activity){}
string describtion() const {
return Pet::name() + "likes to" + favoriteActivity;
}
};
void describtion(Pet *p)
{
string aa = (*p).describtion();
cout << aa.c_str() << endl;
}
int main()
{
Pet *p = new Pet("Alfred");
Dog *g = new Dog("full", "sleep");
describtion(p);
describtion(g);
getchar();
return 0;
}