死亡钻石和范围解析算子（c ++）

本文介绍了死亡钻石和范围解析算子（c ++）的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 29岁程序员，3月因学历无情被辞！ 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 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)(); // errorThe 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(); 这篇关于死亡钻石和范围解析算子（c ++）的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！ 上岸，阿里云！