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68

answers:

2

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?

+3  A: 

The 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.

Tyler McHenry
Your answer is correct. From my research if Derived extends BaseA and BaseB, the object is laid out in memory as |BaseA|BaseB|Derived|. Thus the pointer is pointing to the start of BaseA so casting Derived to BaseB will have you reading BaseA's members.
Chazz
+3  A: 

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*).

James McNellis