死亡钻石和范围解析算子（c ++） [英] Diamond of death and Scope resolution operator (c++)

问题描述

我有这个代码（钻石问题）：

  #include< iostream> 
 using namespace std; 
 
 struct Top 
 {
 void print（）{cout<< Top :: print（）<< ENDL; } 
}; 
 
 struct右：Top 
 {
 void print（）{cout<< Right :: print（）<< ENDL; } 
}; 
 
 struct Left：Top 
 {
 void print（）{cout<< Left :: print（）<< ENDL; } 
}; 
 
 struct Bottom：Right，Left {}; 
 
 int main（）
 {
底部b; 
 b.Right :: Top :: print（）; 
} 
  

我想拨打 print（） in Top class。

当我尝试编译它时，我收到错误：'Top'是这一行中'Bottom'的模糊基础： b.Right :: Top :: print（）;
为什么它含糊不清？我明确指出我希望 Top 来自右而不是来自左。

我不想知道怎么做，是的，它可以用引用，虚拟继承等来完成。我只是想知道为什么是 b.Right :: Top :: print（）; 不明确。

解决方案

< blockquote>

为什么它不明确？我明确指出我希望 Top 来自右而不是来自左。

这是你的意图，但事实并非如此。 Right :: Top :: print（）明确命名要调用的成员函数，即& Top :: print 。但它没有指定 b 的哪个子对象我们正在调用该成员函数。您的代码在概念上等同于：

  auto print =& Bottom :: Right :: Top :: print; // ok 
（b。* print）（）; //错误
  

选择打印的部分是明确的。这是从 b 到 Top 的隐式转换，这是不明确的。您必须通过以下操作明确消除您进入的方向的歧义：

  static_cast< Right&> （b）中.TOP ::打印（）; 
  

I have this code (diamond problem):

#include <iostream>
using namespace std;

struct Top
{
    void print() { cout << "Top::print()" << endl; }
};

struct Right : Top 
{
    void print() { cout << "Right::print()" << endl; }
};

struct Left : Top 
{
    void print() { cout << "Left::print()" << endl; }
};

struct Bottom: Right, Left{};

int main()
{
    Bottom b;
    b.Right::Top::print();
}

I want to call print() in Top class.

When I try to compile it I get error: 'Top' is an ambiguous base of 'Bottom' on this line: b.Right::Top::print(); Why is it ambiguous? I explicitly specified that I want Top from Right and not from Left.

I don't want to know HOW to do it, yes it can be done with references, virtual inheritance, etc. I just want to know why is b.Right::Top::print(); ambiguous.

解决方案

Why is it ambiguous? I explicitly specified that I want Top from Right and not from Left.

That was your intent, but that's not what actually happens. Right::Top::print() explicitly names the member function that you want to call, which is &Top::print. But it does not specify on which subobject of b we are calling that member function on. Your code is equivalent conceptually to:

auto print = &Bottom::Right::Top::print;  // ok
(b.*print)();                             // error

The part that selects print is unambiguous. It's the implicit conversion from b to Top that's ambiguous. You'd have to explicitly disambiguate which direction you're going in, by doing something like:

static_cast<Right&>(b).Top::print();

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