Reputation: 43234
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?
Upvotes: 5
Views: 3479
Reputation: 1777
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.
Upvotes: 0
Reputation: 153909
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.
Upvotes: 7
Reputation:
non-constant lvalue reference (B*&) cannot bind to a unrelated type (A*).
Upvotes: 3
Reputation: 76519
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.
Upvotes: -1