从派生**到基数**的转换-解网

问：

我正在阅读这篇文章，不幸的是，我无法深入理解为什么编译器不允许从 Derived** 转换为 Base**。我也看到了这一点，它提供的信息只比 parashift.com 的链接多。

编辑：

让我们逐行分析这段代码：

   Car   car;
   Car*  carPtr = &car;
   Car** carPtrPtr = &carPtr;
   //MyComment: Until now there is no problem!

   Vehicle** vehiclePtrPtr = carPtrPtr;  // This is an error in C++
   //MyComment: Here compiler gives me an error! And I try to understand why. 
   //MyComment: Let us consider that it was allowed. So what?? Let's go ahead!

   NuclearSubmarine  sub;
   NuclearSubmarine* subPtr = &sub;
   //MyComment: this two line are OK too!

   *vehiclePtrPtr = subPtr;

   //MyComment: the important part comes here... *vehiclePtrPtr is a pointer to
   //MyComment: a vehicle, particularly in our case it points to a Car object.
   //MyComment: Now when I assign to the pointer to the Car object *vehiclePtrPtr,
   //MyComment: a pointer to NuclearSubmarine, then it should just point to the
   //MyComment: NuclearSubmarine object as it is indeed a pointer to a Vehicle,
   //MyComment: isn't it? Where is my fault? Where I am wrong?

   // This last line would have caused carPtr to point to sub!
   carPtr->openGasCap();  // This might call fireNuclearMissle()!

C 继承针 c++-faq 指针转换

理解 FAQ 示例的关键是，在实践中，多态对象包含指向函数指针表的隐藏指针，该指针指向其虚拟函数。因此，如果您可以指向 A 而不是，那么调用虚拟函数可能会调用潜艇的导弹发射。造成灾难性后果。NuclearSubmarine*CarNuclearSubmarineCar

0赞 David Schwartz 11/6/2011

当然，没有 NuclearSubmarine：：openGasCap 函数，因此如果允许建议的转换，此代码将调用不存在的函数，而不会出现强制转换或警告。

答：

11赞 David Schwartz 11/6/2011 #1

不乏无意义的错误，这将允许：

class Flutist : public Musician
...

class Pianist : public Musician
...

void VeryBad(Flutist **f, Pianist **p)
{
 Musician **m1=f;
 Musician **m2=p;
 *m1=*m2; // Oh no! **f is supposed to be a Flutist and it's a Pianist!
}

下面是一个完整的工作示例：

#include <stdio.h>

class Musician
{
 public:
 Musician(void) { ; }
 virtual void Play(void)=0;
};

class Pianist : public Musician
{
 public:
 Pianist(void) { ; }
 virtual void Play(void) { printf("The piano blares\n"); }
};

class Flutist : public Musician
{
 public:
 Flutist(void) { ; }
 virtual void Play(void) { printf("The flute sounds.\n"); }
};

void VeryBad(Flutist **f, Pianist **p)
{
 Musician **m1=f;
 Musician **m2=p;
 *m1=*m2; // Oh no! **f is supposed to be a Flutist and it's a Pianist!
}

int main(void)
{
 Flutist *f=new Flutist();
 Pianist *p=new Pianist();
 VeryBad(&f, &p);
 printf("Mom is asleep, but flute playing wont bother her.\n");
 f->Play(); // Since f is a Flutist* this can't possibly play piano, can it?
}

这是它的实际应用：

$ g++ -fpermissive verybad.cpp -o verybad
verybad.cpp: In function void VeryBad(Flutist**, Pianist**):
verybad.cpp:26:20: warning: invalid conversion from Flutist** to Musician** [-fpermissive]
verybad.cpp:27:20: warning: invalid conversion from Pianist** to Musician** [-fpermissive]
$ ./verybad 
Mom is asleep, but flute playing wont bother her.
The piano blares

我不明白这个例子。“// 既然 p 是长笛演奏家*，这不可能弹钢琴，对吧？”，但是一个，它应该弹钢琴，它在弹钢琴，问题出在哪里？在分配了一个，但如果这导致了一个问题，它要么没有在示例中演示，要么我完全错过了这一点pPianist*VeryBadFlutist*Pianist*

1赞 463035818_is_not_an_ai 5/9/2017

没关系。我不确定这个例子或我的理解是否有问题。我在其他地方读过一点，现在或多或少很清楚了。你的例子真的有很大帮助。

23赞 fredoverflow 11/6/2011 #2

这基本上与一碗香蕉不是一碗水果的原因相同。如果一碗香蕉是一碗水果，你可以把一个苹果放进碗里，它就不再是一碗香蕉了。

只要你只检查碗，转换是无害的。但是一旦你开始修改它，转换就会变得不安全。这是要牢记的关键点。（这就是为什么不可变的 Scala 集合实际上允许转换，但可变集合禁止转换的确切原因。

与您的示例相同。如果有从到的转换，你可以放置一个指向苹果的指针，如果类型系统承诺只能存在一个指向香蕉的指针。繁荣！Derived**Base**

从派生到基数的转换

Conversion from Derived to Base

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从派生**到基数**的转换

Conversion from Derived** to Base**

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从派生到基数的转换

Conversion from Derived to Base