问题
I'm trying to get a better understanding how virtual inheritance works in practice (that is, not according to the standard, but in an actual implementation like g++
). The actual question is at the bottom, in bold face.
So, I built myself a inheritance graph, which has among other things, these simple types:
struct A {
unsigned a;
unsigned long long u;
A() : a(0xAAAAAAAA), u(0x1111111111111111ull) {}
virtual ~A() {}
};
struct B : virtual A {
unsigned b;
B() : b(0xBBBBBBBB) {
a = 0xABABABAB;
}
};
(In the whole hierarchy I also have a C: virtual A
and BC: B,C
, so that the virtual inheritance makes sense.)
I wrote a couple of functions to dump the layout of instances, taking the vtable pointer and printing the first 6 8-byte values (arbitrary to fit on screen), and then dump the actual memory of the object. This looks something like that:
Dumping an A
object:
actual A object of size 24 at location 0x936010
vtable expected at 0x402310 {
401036, 401068, 434232, 0, 0, 0,
}
1023400000000000aaaaaaaa000000001111111111111111
[--vtable ptr--]
Dumping a B
object and where the A
object is located, which is indicated by printing a lot A
s at the respective position.
actual B object of size 40 at location 0x936030
vtable expected at 0x4022b8 {
4012d2, 40133c, fffffff0, fffffff0, 4023c0, 4012c8,
}
b822400000000000bbbbbbbb00000000e022400000000000abababab000000001111111111111111
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA (offset: 16)
As you can see, the A
part of B
is located at an offset of 16
bytes to the start of the B
object (which could be different if I had instantiated BC
and dyn-casted it to a B*
!).
I would have expected the 16
(or at least a 2
, due to alignment) to show up somewhere in the table, because the program has to look up the actual location (offset) of A
at runtime. So, how does the layout really look like?
Edit: The dump is done by calling dump
and dumpPositions
:
using std::cout;
using std::endl;
template <typename FROM, typename TO, typename STR> void dumpPositions(FROM const* x, STR name) {
uintptr_t const offset {reinterpret_cast<uintptr_t>(dynamic_cast<TO const*>(x)) - reinterpret_cast<uintptr_t>(x)};
for (unsigned i = 0; i < sizeof(FROM); i++) {
if (offset <= i && i < offset+sizeof(TO))
cout << name << name;
else
cout << " ";
}
cout << " (offset: " << offset << ")";
cout << endl;
}
template <typename T> void hexDump(T const* x, size_t const length, bool const comma = false) {
for (unsigned i = 0; i < length; i++) {
T const& value {static_cast<T const&>(x[i])};
cout.width(sizeof(T)*2);
if (sizeof(T) > 1)
cout.fill(' ');
else
cout.fill('0');
cout << std::hex << std::right << (unsigned)value << std::dec;
if (comma)
cout << ",";
}
cout << endl;
}
template <typename FROM, typename STR> void dump(FROM const* x, STR name) {
cout << name << " object of size " << sizeof(FROM) << " at location " << x << endl;
uintptr_t const* const* const vtable {reinterpret_cast<uintptr_t const* const*>(x)};
cout << "vtable expected at " << reinterpret_cast<void const*>(*vtable) << " {" << endl;
hexDump(*vtable,6,true);
cout << "}" << endl;
hexDump(reinterpret_cast<unsigned char const*>(x),sizeof(FROM));
}
回答1:
The answer is actually documented here, in the Itanium ABI. In particular section 2.5 contains the layout of the virtual table.
来源:https://stackoverflow.com/questions/9947805/how-does-the-virtual-inheritance-table-work-in-g