什么是对象切片?

c++在将一个派生类转换为基类的过程中,派生类的一部分将被基类接收不到,只能留下基类大小的对象。

传值调用的切片:

#include<iostream>
using namespace std;
class Base
{
public:
    virtual void func1()
    {
        cout << "Base::func1" << endl;
    }
    virtual void func2()
    {
        cout << "Base::func2" << endl;
    }
private:
    int a;
};
class Derive :public Base
{
public:
    virtual void func1()
    {
        cout << "Derive::func1" << endl;
    }
    virtual void func3()
    {
        cout << "Derive::func3" << endl;
    }
    virtual void func4()
    {
        cout << "Derive::func4" << endl;
    }
private:
    int b;
};
typedef void(*FUNC) ();
void PrintVTable(int* VTable)
{
    cout << " 虚表地址>" << VTable << endl;
    for (int i = 0; VTable[i] != 0; ++i)
    {
        printf(" 第%d个虚函数地址 :0X%x,->", i, VTable[i]);
        FUNC f = (FUNC)VTable[i];
        f();
    }
    cout << endl;
}
void function(Base b)
{
    b.func1();
}
void Test1()
{
    Base b1;
    Derive d1;
    int* VTable1 = (int*)(*(int*)&b1);
    int* VTable2 = (int*)(*(int*)&d1);
    PrintVTable(VTable1);
    PrintVTable(VTable2);
    function(d1);
}
int main()
{
    Test1();
    system("pause");
}


再拷贝的过程中发生了切片,在调用构造函数的时候初始化VPTR指向基类的VTABLE,并且只拷贝了对象的基类部分,所以最后就变成了一个基类的对象。

如果要防止这种现象的发生,只需把基类定义成纯虚函数就可以了。

将派生类的指针传递给基类

#include<iostream>
using namespace std;
class Base
{
public:
    virtual void func1()
    {
        cout << "Base::func1" << endl;
    }
    virtual void func2()
    {
        cout << "Base::func2" << endl;
    }
private:
    int a;
};
class Derive :public Base
{
public:
    virtual void func1()
    {
        cout << "Derive::func1" << endl;
    }
    virtual void func3()
    {
        cout << "Derive::func3" << endl;
    }
    virtual void func4()
    {
        cout << "Derive::func4" << endl;
    }
private:
    int b;
};
typedef void(*FUNC) ();
void PrintVTable(int* VTable)
{
    cout << " 虚表地址>" << VTable << endl;
    for (int i = 0; VTable[i] != 0; ++i)
    {
        printf(" 第%d个虚函数地址 :0X%x,->", i, VTable[i]);
        FUNC f = (FUNC)VTable[i];
        f();
    }
    cout << endl;
}
void function(Base* b)
{
    b->func1();
    b->func2();
}
void Test1()
{
    Base b1;
    Derive d1;
    int* VTable1 = (int*)(*(int*)&b1);
    int* VTable2 = (int*)(*(int*)&d1);
    PrintVTable(VTable1);
    PrintVTable(VTable2);
    function((Base*)&d1);
}
int main()
{
    Test1();
    system("pause");
}


解释:

在将派生类的指针传给基类之后,就将派生类的大小看作是积累的大小,所以就只能访问基类的大小,由于没有发生构造函数,所以VPTR还是派生类的。

Last modification:April 9, 2020
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