为什么没有static_cast会失败？

Question

Compiling f works, but compiling g fails with an error.

Why does this happen?

class A {
public:
  A() {}
};

class B : public A {
public:
  B() {}
};

void f() {
  A* a = new A();
  B* b = static_cast<B*>(a);
}

void g() {
  A* a = new A();
  B* b = a;
}

Answer 1

A static_cast forces a conversion that is potentially unsafe.

B* b = static_cast<B*>(a);

This would be valid if a pointed to an A object that actually was the base class sub-object of a B object, however it doesn't. The cast forces the conversion.

B* b = a;

There is no cast here and there is (correctly) no implicit conversion allowed from base class pointer to derived class pointer. A pointer to a derived class can always be converted to a pointer to a base class because a derived class object always contains a base class sub-object but not every base class instance is a sub-object of a particular derived class type.

Answer 2

Well, yeah. Doing:

B* b = new A();

Is unsafe. You end up with a B pointer to an A object; you never construct the B portion of the object; your object is "sliced".

On the other hand...

A* a = new B();

...would be fine.

Answer 3

You are trying to convert a pointer from A* to B*. I am not sure what you are trying to achieve. But since B* is derived from A* and not the other way around this is not valid. Maybe you want to do something like this:

int main()
{
///The above code compiles while if I replace above two line in main with below assignment it gives error.
     A *a=new A();
    A * b=new B();

}

Answer 4

Yes, it does give an error if you want to assign a base class to a derived class pointer type. No, it doesn't give an error if you explicitly cast the pointer type, because in C++ you are allowed to shoot yourself in the foot if you so desire.

What exactly is baffling you, or what did you expect to achieve with your code?

Answer 5

A base class cannot be implicitly converted to a derived class. Just consider this

class A {
  public: int x;
};
class B : public A {
  public: int y;
};
B* b = new A; // assume it works, so 4 bytes is allocated and initialized.
b->y;         // bam! accessing unallocated  region.