为什么选择C++中的dynamic_cast

Question

考虑以下代码:

#include <iostream>

using namespace std;

class Base{
    int i;
    public:
    virtual bool baseTrue() {return true;}
    Base(int i) {this->i=i;}
    int get_i() {return i;}
    };

class Derived : public Base{
    int j;
    public:
    Derived(int i,int j) : Base(i) {this->j=j;}
    int get_j() {return j;}
    };

int main()
{
    Base *bp;
    Derived *pd,DOb(5,10);

    bp = &DOb;

    //We are trying to cast base class pointer to derived class pointer
    cout << bp->get_i() << endl;
    cout << ((Derived *)bp)->get_j() << endl;**//HERE1**

    pd=dynamic_cast<Derived*> (bp); **//HERE2**
    // If base class is not polymorphic
    //throw error
    //error: cannot dynamic_cast `bp' (of type `class Base*') to
    //type `class Derived*' (source type is not polymorphic)

    cout << pd->get_j() << endl;**//HERE2**

    //Now we try to cast derived Class Pointer to base Class Pointer

    Base *pb;
    Derived *dp,Dbo(50,100);
    dp = &Dbo;


    cout << ((Base *)dp)->get_i() << endl;**//HERE3**
    //cout << ((Base *)dp)->get_j() << endl;
    //throws error Test.cpp:42: error: 'class Base' has no member named 'get_j'

    pb =  dynamic_cast<Base * > (dp); **//HERE4**
    cout << pb->get_i() << endl; **//HERE4**
    //cout << pb->get_j() << endl;
    //throws error Test.cpp:47: error: 'class Base' has no member named 'get_j'


    return 0;
    }

输出

Gaurav@Gaurav-PC /cygdrive/d/Glaswegian/CPP/Test
$ ./Test
5
10
10
50
50

我的投射方式(Line HERE1和HERE2)和(HERE3&HERE4),这两者有什么区别？两者都产生相同的输出,那么为什么要选择dynamic_cast

Answer 1

dynamic_cast 是"安全的",因为它会抛出一个异常,或者在你做"坏"的事情时返回NULL(或者,正如Nawaz所说,它不会编译,因为类型足够糟糕,编译器可以看到它出错)

的(Derived *)...形式将作用类似reinterpret_cast<Derived *>(...),这是"不安全" -它将一个指针简单地转换为其它指针类型,是否能产生一个有意义的结果或没有.如果表现"糟糕",那就是你的问题.

你可以这样做:

int x = 4711;

Derived *dp = (Derived *)x; 
cout << dp->get_j(); 

编译器可能会对整数的大小抱怨,但否则,它会编译代码.它很可能根本不会运行,但如果确实如此,结果可能没什么"有用".