为什么必须由派生类最多的类构造虚拟基类？ [英] Why must virtual base classes be constructed by the most derived class?

问题描述

以下代码无法编译：

class A {
public:
    A(int) {}
};

class B: virtual public A {
public:
    B(): A(0) {}
};

// most derived class
class C: public B {
public:
    C() {} // wrong!!!
};

如果我打电话给 A 的构造函数 C 的构造函数初始化列表，即：

If I call A's constructor in C's constructor initialization list, that is:

// most derived class
class C: public B {
public:
    C(): A(0) {} // OK!!!
};

它确实有效。

显然，原因是虚拟基类必须始终由大多数派生类构造。

Apparently, the reason is because virtual base classes must always be constructed by the most derived classes.

我不明白这个限制背后的原因。

I don't understand the reason behind this limitation.

推荐答案

因为它避免了这个：

class A {
public:
    A(int) {}
};

class B0: virtual public A {
public:
    B0(): A(0) {}
};

class B1: virtual public A {
public:
    B1(): A(1) {}
};

class C: public B0, public B1 {
public:
    C() {} // How is A constructed? A(0) from B0 or A(1) from B1?
};

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