c++ - 在构造期间隐式转换为基类

Question

12.7/3中给出了下面的例子:

struct A { };
struct B : virtual A { };
struct C : B { };
struct D : virtual A { D(A*); };
struct X { X(A*); };
struct E : C, D, X {
    E() : D(this), // undefined: upcast from E* to A*
                   // might use path E* → D* → A*
                   // but D is not constructed
                   // D((C*)this), // defined:
                   // E* → C* defined because E() has started
                   // and C* → A* defined because
                   // C fully constructed
    X(this) {      // defined: upon construction of X,
                   // C/B/D/A sublattice is fully constructed
    }
};

示例规则如下:

To explicitly or implicitly convert a pointer (a glvalue) referring to an object of class X to a pointer (reference) to a direct or indirect base class B of X, the construction of X and the construction of all of its direct or indirect bases that directly or indirectly derive from B shall have started and the destruction of these classes shall not have completed, otherwise the conversion results in undefined behavior.

在标准提供的情况下，我们有 A 派生类的列表是 {B, D, C, E}。 A、C 和B 的构建已经完成。但是 D 的构造在我们需要向上转换使用 E* → D* → A* 的时候已经开始了。那么为什么实际行为未定义？

Answer 1

到 A* 的转换是一个参数评估，它必须在结果传递给 D 构造函数之前完成。所以D的构建还没有开始。作为反例，E 的构造已经此时开始，尚未完成。