conv.h
class Base
{
public:
void foo();
};
class Derived: public Base
{
public:
void bar();
};
class A {};
class B
{
public:
void koko();
};
Run Code Online (Sandbox Code Playgroud)
conv.cpp
void Base::foo()
{
cout<<"stamm";
}
void Derived::bar()
{
cout<<"bar shoudn't work"<<endl;
}
void B::koko()
{
cout<<"koko shoudn't work"<<endl;
}
Run Code Online (Sandbox Code Playgroud)
main.cpp中
#include "conv.h"
#include <iostream>
int main()
{
Base * a = new Base;
Derived * b = static_cast<Derived*>(a);
b->bar();
Derived * c = reinterpret_cast<Derived*>(a);
c->bar();
A* s1 = new A;
B* s2 = reinterpret_cast<B*>(s1);
s2->koko();
}
Run Code Online (Sandbox Code Playgroud)
输出:
bar shoudn't work
bar shoudn't work
koko shoudn't work
Run Code Online (Sandbox Code Playgroud)
为什么在运行时调用方法栏成功,尽管我创建了一个未派生的Base类?它甚至可以用于两种类型的转换(静态和重新解释转换).
与上述相同的问题,但与不相关的类(A和B)?