Question

我正在尝试第一次实施5条规则。阅读了很多有关最佳做法的建议后，我得出了一个解决方案，其中复制/移动赋值运算符似乎有些冲突。

这是我的代码。

#include <vector>   
#include <memory>

template<class T> class DirectedGraph {
public:
    std::vector<T> nodes;
    DirectedGraph() {}
    DirectedGraph(std::size_t n) : nodes(n, T()) {}
    // ... Additional methods ....
};

template<class T>
DirectedGraph<T> Clone(DirectedGraph<T> graph) {
    auto clone = DirectedGraph<T>();
    clone.nodes = graph.nodes;
    return clone;
}

template<class T> class UndirectedGraph
{
    using TDirectedG = DirectedGraph<T>;
    using TUndirectedG = UndirectedGraph<T>;

    std::size_t numberOfEdges;
    std::unique_ptr<TDirectedG> directedGraph;
public:
    UndirectedGraph(std::size_t n)
        : directedGraph(std::make_unique<TDirectedG>(n))
        , numberOfEdges(0) {}

    UndirectedGraph(TUndirectedG&& other) {
        this->numberOfEdges = other.numberOfEdges;
        this->directedGraph = std::move(other.directedGraph);
    }

    UndirectedGraph(const TUndirectedG& other) {
        this->numberOfEdges = other.numberOfEdges;
        this->directedGraph = std::make_unique<TDirectedG>
            (Clone<T>(*other.directedGraph));
    }

    friend void swap(TUndirectedG& first, TUndirectedG& second) {
        using std::swap;
        swap(first.numberOfEdges, second.numberOfEdges);
        swap(first.directedGraph, second.directedGraph);
    }

    TUndirectedG& operator=(TUndirectedG other) {
        swap(*this, other);
        return *this;
    }

    TUndirectedG& operator=(TUndirectedG&& other) {
        swap(*this, other);
        return *this;
    }

    ~UndirectedGraph() {}
};

int main()
{
    UndirectedGraph<int> graph(10);
    auto copyGraph = UndirectedGraph<int>(graph);
    auto newGraph = UndirectedGraph<int>(3);
    newGraph = graph;            // This works.
    newGraph = std::move(graph); // Error here!!!
    return 0;
}

我从here获得的大部分建议，我实现了副本分配operator=来接受 value 的参数。我认为这可能是个问题，但我不明白为什么。

此外，如果有人指出我的复制/移动ctor /分配是否以正确的方式实施，我将不胜感激。

Answer 1

您应该具有：

TUndirectedG& operator=(const TUndirectedG&);
TUndirectedG& operator=(TUndirectedG&&);

或

TUndirectedG& operator=(TUndirectedG);

同时拥有

TUndirectedG& operator=(TUndirectedG);   // Lead to ambiguous call
TUndirectedG& operator=(TUndirectedG&&); // Lead to ambiguous call

将导致带有右值的模糊调用。

c ++：实现移动分配时，operator =不明确

1 个答案: