static_cast和对指针的引用

Question

Can anyone tell me why this doesn't compile:

struct A { };
struct B : public A { };

int main()
{
  B b;
  A* a = &b;
  B* &b1 = static_cast<B*&>(a);
  return 0;
}

Now, if you replace the static cast with:

B* b1 = static_cast<B*>(a);

then it does compile.

Edit: It is obvious that the compiler treats A* and B* as independent types, otherwise this would work. The question is more about why is that desirable?

Answer 1

B is derived from A , but B* isn't derived from A* . A pointer to a B is not a pointer to an A , it can only be converted to one. But the types remain distinct (and the conversion can, and often will, change the value of the pointer). A B*& can only refer to a B* , not to any other pointer type.

Answer 2

非常量左值引用（B *＆）不能绑定到不相关的类型（A *）。

Answer 3

You are trying to cast an A* to a B* . This is the wrong way around and not very useful. You probably want to store a pointer to derived in a pointer to base, which is useful and doesn't even need a cast.

I suppose a dynamic_cast might work here, but the result is implementation defined if I'm not mistaken.

Answer 4

Handling of references is something the compiler does for you, there should be no need to cast to reference.

If we refactor the code to:

B b;
A* a = &b;
B* b_ptr = static_cast<B*>(a);
B*& p1 = b_ptr;

It will compile.