If you have a void* pointer to Derived class that inherits from both BaseA
and BaseB
, how does the compiler cast the void*
pointer to BaseA*
(or BaseB*
) without knowing that the void*
pointer is of type Derived
?
views:
68answers:
2The compiler doesn't cast the void*
pointer to anything -- you, the programmer, do.
In order to do anything useful with a void*
pointer, you need to explicitly cast it to a non-void*
pointer, and if you're wrong about what type the pointer actually points to, you enter Undefined Behavior City.
It doesn't. The only guarantee when casting to and from a void*
using a static_cast
is:
A value of type pointer to object converted to "pointer to
cv void
" and back to the original pointer type will have its original value (C++03 §5.2.9/10).
For example, the following code is incorrect because the void*
is cast to a type other than the original pointer type (the cast sequence is B1*
-> void*
-> B2*
):
struct B1 { int i; };
struct B2 { int j; };
struct D : B1, B2 { };
D x;
B1* b1ptr = &x;
void* voidptr = b1ptr;
B2* b2ptr = static_cast<B2*>(voidptr);
Attempting to use b2ptr
here would result in undefined behavior. The only type to which you can safely cast voidptr
is B1*
, since that is the type from which the void*
was obtained (well, or to a char*
, since anything can be accessed via a char*
).