从派生**到基础**的转换

Question

我正在阅读此，不幸的是无法理解深入了解为什么编译器不允许从 Derived** 到 Base** 的转换。我还看到 this 没有给出比 parashift.com 的链接有更多信息。

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 = ⊂
   //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()!

Answer 1

这基本上和一碗香蕉不是一碗水果的原因是一样的。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。

只要您只检查碗，这种转换是无害的。但一旦您开始修改它，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。如果存在从 Derived** 到 Base** 的转换，则您可以放置​​一个指向苹果的指针，而类型系统承诺只能存在指向香蕉的指针。繁荣！

Answer 2

这将允许不乏无意义的错误：

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!
}

这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?
}

这里是它的实际应用：

$ 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

Answer 3

不允许 Vehicle**vehiclePtrPtr = carPtrPtr;，因为它是 Derived** 到 Base** 的转换，这是不允许的。

您的示例中说明了不允许的原因。

   Car   car;
   Car*  carPtr = &car;
   Car** carPtrPtr = &carPtr; 

这样 carPtrPtr 就指向 Car 的指针。

核潜艇；
NuclearSubmarine* subPtr = ⊂

这也是合法的，但如果您可以这样做，

Vehicle** vehiclePtrPtr = carPtrPtr;

您可能会意外地

*vehiclePtrPtr = subPtr;

这样做，因为 *vehiclePtrPtr 是指向 Car 的指针。在最后一行中，您为其分配一个指向 Sub 的指针。因此，您现在可以在 Car 类型的对象上调用派生类 Sub 中定义的方法，但具有未定义的行为。