在构造函数中使用object-expression调用虚函数

问题描述

代码：

 c  
 
否则， x 是包含 x 的（唯一）对象的完整对象。
 
 

  
If x is a complete object, then x is the complete object of x.
  
Otherwise, the complete object of x is the complete object of the (unique) object that contains x.
  
实质上，你引用的句子使得 static_cast< C *>（this） - > foo（）;  in  B 的构造函数未定义的行为，即使正在构造的完整对象是 C 。标准实际上提供了一个很好的例子：
In essence, the sentence you cited makes doing static_cast<C*>(this)->foo(); in B's constructor undefined behavior in your code, even if the complete object being constructed is a C. The standard actually provides a pretty good example here:
struct V {
    virtual void f();
    virtual void g();
};
struct A : virtual V {
    virtual void f();
};
struct B : virtual V {
    virtual void g();
    B(V*, A*);
};

struct D : A, B {
    virtual void f();
    virtual void g();
    D() : B((A*)this, this) { }
};

B::B(V* v, A* a) {
    f();    // calls V::f, not A::f
    g();    // calls B::g, not D::g
    v->g(); // v is base of B, the call is well-defined, calls B::g
    a->f(); // undefined behavior, a’s type not a base of B
}
可以已经：Ideone的编译器（GCC）实际调用 V :: f（）在 a-> f（）; 行，即使指针指的是完全构造的 A 子对象。
In fact, you can already see the undefined behavior show up in this example if you run it: Ideone's compiler (GCC) actually calls V::f() on the a->f(); line, even though the pointer is referring to a fully constructed A subobject.
                        
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