c++ - 在 C++ 中使用不正确的向下转换访问基类成员

Question

下面的 C++ 代码是否正确？

struct Base { int x; };
struct Derived : Base { int y; }
Base * b = new Base;
Derived * d = static_cast<Derived *>(b);
//below we access only d->x, but not d->y
std::cout << d->x;

如果不是，那到底是哪里出了问题？ C++ 标准对此有何评论？至少我还没有看到它崩溃过。

Answer 1

这在 [expr.static.cast]/11 中相当简单(强调我的):

A prvalue of type “pointer to cv1 B”, where B is a class type, can be converted to a prvalue of type “pointer to cv2 D”, where D is a class derived from B, if cv2 is the same cv-qualification as, or greater cv-qualification than, cv1. If B is a virtual base class of D or a base class of a virtual base class of D, or if no valid standard conversion from “pointer to D” to “pointer to B” exists ([conv.ptr]), the program is ill-formed. The null pointer value is converted to the null pointer value of the destination type. If the prvalue of type “pointer to cv1 B” points to a B that is actually a subobject of an object of type D, the resulting pointer points to the enclosing object of type D. Otherwise, the behavior is undefined.

您没有 Derived 的子对象，因此这是未定义的行为。

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请注意，左值或 xvalue 指针没有特殊情况，并且/8 提到操作数经历了左值到右值的转换。