虚拟基类和初始化列表 [英] virtual base class and initialization lists

问题描述

可能重复：

gcc c ++虚拟继承问题

p>

我想知道当使用来自虚拟基类的多重继承时，编译器如何处理不同的初始化值。考虑臭名昭着的钻石的恐惧继承计划：

 基本
 / \ 
 / \ 
 D1 D2 
 \ / 
 \ / 
加入
  

$ b b

为了避免在 Join 中有 Base 的两个副本，我使用虚拟继承 D1 和 D2 （参见例如此处）。现在，让 Base 不是抽象的，但有一个成员字段，它在其构造函数中初始化：

  class Base {
 public：
 Base（int x_）{x = x_;}; 
 virtual〜Base（）{}; 
 
 public：
 int x; 
}; 
 
 class D1：public virtual Base {
 public：
 D1（）：Base（1）{}; 
 virtual〜D1（）{}; 
}; 
 
 class D2：public virtual Base {
 public：
 D2（）：Base（2）{}; 
 virtual〜D2（）{}; 
}; 
 
类加入：public D1，public D2 {
 public：
 Join（）{}; 
〜Join（）{}; 
}; 
 
 int main（）
 {
加入j; 
 cout<< j.x<< endl; 
 
 return 0; 
} 
  

输出是1,2还是编译器依赖？ / p>

解决方案

它不编译。虚拟基址由最小的导出类 Join 初始化。 加入未显式初始化 Base ，但 Base

[它也不会编译，因为类的定义需要以; 但我认为这是一个错字。 main 也应该返回 int ，我假设 jx 是 j-> x 的拼写错误。]

Possible Duplicate:
gcc c++ virtual inheritance problem

Hi all,

I'm wondering about how the compiler would handle different initialization values when using multiple inheritance from a virtual base class. Consider the notorious 'diamond of dread' inheritance scheme:

     Base
     / \
    /   \
  D1     D2
    \   /
     \ /
     Join

In order to avoid having two copies of Base in Join, I use virtual inheritance for D1 and D2 (see e.g. here). Now, lets say Base is not abstract, but has a member field, which is initialized in its constructor:

class Base { 
public:
    Base(int x_) {x = x_;};
    virtual ~Base(){};

public:
    int x;
};

class D1 : public virtual Base {
public:
    D1() : Base(1) {};
    virtual ~D1(){};
};

class D2 : public virtual Base {
public:
    D2() : Base(2) {};
    virtual ~D2(){};
};

class Join : public D1, public D2 {
public:
    Join(){};
    ~Join(){};
};

int main()
{
   Join j;
   cout << j.x << endl;

   return 0;
}

Will the output be 1, 2, or is it compiler-dependent?

解决方案

It shoudn't compile. Virtual bases are initialized by the most derived class which is Join. Join doesn't explicitly initialize Base but Base has no accessible default constructor.

[It also won't compiler because definitions of classes need to be terminated with a ; but I've assumed that this is a typo. main should also return int and I've assumed that j.x is a typo for j->x.]

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