问题描述
struct B { int i; };
struct D1 : virtual B {};
struct D2 : B {}; // <-- not virtual
struct DD : D1, D2 {};
编译器需要 D2 virtual ：
DD d;
d.i = 0; // error: request for member `i' is ambiguous
我不明白的是，您提示编译器 B 是 virtual > D1 ），为什么仍然 i 不明确？
What I don't understand is, once you have prompted compiler that B is virtual with respect to DD (via D1) then why it still i is ambiguous ?
（如果我的内存正确，老的VC ++（在2006年）足够能够用单个 virtual 继承）来证明这一点。
(If my memory serves correct, the older VC++ (in 2006), was capable enough to make out this just with single virtual inheritance)
推荐答案
B相对于DD不是虚拟的 - 它相对于D1是虚拟的。在D2被创建时，它包含一个B的完整副本。所以现在DD有两个B的实现：一个作为D2的一部分，一个在结束（由D1指向）。有两个副本 i ，使用它确实是不明确的。
B is not virtual with respect to DD - it is virtual with respect to D1. At the time D2 is created, it contains a full copy of B. So now DD has two implementations of B: one as part of D2, and one at the end (pointed by D1). And having two copies of i, using it is indeed ambiguous.
D2也使用虚拟继承，而不是包含B的副本，它将包含指向D1的实例的指针，D1也指向，并且DD将仅包含B的一个实例。
Had D2 also used virtual inheritance, instead of containing a copy of B, it would have contained a pointer to the instance of B that D1 is also pointing at, and DD would have contained only one instance of B.
我会尝试说明内存布局，希望这出来正确...：
I'll try to illustrate the memory layouts, hope this comes out right...:
你的情况下，一个虚拟继承和一个非虚拟 -
Your case, with one virtual inheritance and one non-virtual -
| D1 | D2 + B | B |
+--+-------+----------+---------+
| vptr to B ^
+-----------------------|
D1和D2实际上都继承了 -
Having both D1 and D2 inherit virtually -
| D1 | D2 | B |
+--+-----+---+----+-------+
| | ^
+---------+---------|
这篇关于为什么单个虚拟继承不足以解决可怕的钻石问题？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！