为什么在c++中选择dynamic_cast

Why to opt for dynamic_cast in C++

本文关键字：dynamic cast 选择 c++ 为什么      更新时间：2023-10-16

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

我使用(Line HERE1 and HERE2) &(HERE3和;那么，两者的区别是什么呢?那么为什么要使用dynamic_cast

呢?

dynamic_cast是"安全的"，因为当你做一些"不好"的事情时，它会抛出异常或返回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)

(Derived *)...形式的行为与reinterpret_cast<Derived *>(...)类似，这是"不安全的"-它将简单地将一个指针转换为另一个指针类型，无论是否产生有意义的结果。如果它表现得"不好"，那是你的问题。

你可以这样做:

int x = 4711;
Derived *dp = (Derived *)x; 
cout << dp->get_j();

编译器可能会抱怨整数的大小，但除此之外，它将编译代码。它很可能根本不会运行，但即使运行了，结果也可能没什么"有用的"。

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