c++ - 嵌套模板类和全局命名空间中的函数

Question

我正在开发一个模板图数据结构，它是 GraphNode 对象的 STL vector 。我定义了嵌套在 Graph 类中的 GraphNode 类，当我在 Graph 对象 Visual Studio 15 (C++) 报告的重载插入运算符内调用 GraphNode 对象的重载插入运算符时，

(30): warning C4346: 'myGraph<T>::myGraphNode': dependent name is not a type 
(30): note: prefix with 'typename' to indicate a type 
(30): error C2061: syntax error: identifier 'myGraphNode' 
(33): error C2805: binary 'operator <<' has too few parameters 

template <typename T>
ostream& operator<<(ostream& strm, const myGraph<T>::myGraphNode& gn)

将单词 typename 添加到第二个形参

template <typename T>
ostream& operator<<(ostream& strm, typename const myGraph<T>::myGraphNode& gn)

编译器生成以下错误

(49): error C2679: binary '<<': no operator found which takes a right-hand operand of type 'const myGraph<int>::myGraphNode' (or there is no acceptable conversion)

如果我输入 name const .... 或 const typename ...，我会得到同样的错误

为了完整起见，这里对本文的所有代码进行了一些简化。 感谢您的帮助

#include <iostream>
#include <vector>
#include <string>

using namespace std;

typedef unsigned int uint;

template <typename T>
class myGraph {
public:
    class myGraphNode {
    public:
        myGraphNode(T val = T());
        T mData;
    }; // end class myGraphNode

    myGraph();

    uint addGraphNode(T data);
    vector<myGraphNode> mGraphNodes;
}; // end class myGraph


//          myGraphNode
template <typename T>
myGraph<T>::myGraphNode::myGraphNode(T val) : mData(val) {}

template <typename T>
ostream& operator<<(ostream& strm, typename const myGraph<T>::myGraphNode& gn) {
    strm << gn.mData << std::endl;
    return strm;
}


//          myGraph
template <typename T>
myGraph<T>::myGraph() {}

template <typename T>
uint myGraph<T>::addGraphNode(T data) {
    myGraph<T>::myGraphNode node(data);
    mGraphNodes.push_back(node);
}

template <typename T>
ostream& operator<<(ostream& strm, const myGraph<T>& g) {
    for (uint i = 0; i < g.mGraphNodes.size(); ++i)
        cout << g.mGraphNodes[i] << endl;
    return strm;
} // end operator<<(...)

int main()
{
    myGraph<int> g;
    g.addGraphNode(3);
    g.addGraphNode(5);
    cout << g << endl;
    return 0;
}

Answer 1

首先，参数声明的正确语法应该是

template <typename T>
ostream& operator<<(ostream& strm, const typename myGraph<T>::myGraphNode& gn)
//                                 ~~~~~ ~~~~~~~~

引用here了解更多信息。

其次，通过上述声明，当尝试在 operator<< 中调用它时对于 myGraph<T>喜欢 cout << g.mGraphNodes[i] << endl; , T无法推断，因为 non-deduced contexts) :

The nested-name-specifier (everything to the left of the scope resolution operator ::) of a type that was specified using a qualified-id:

这意味着，您必须为其显式指定模板参数，例如

operator<<<T>(strm, g.mGraphNodes[i]);
//        ~~~

但是很丑。对于您的情况，您只需实现 operator<<对于 myGraph<T>喜欢

template <typename T>
ostream& operator<<(ostream& strm, const myGraph<T>& g) {
    for (uint i = 0; i < g.mGraphNodes.size(); ++i)
        cout << g.mGraphNodes[i].mData << endl;
    return strm;
}

顺便说一句:您应该给出 myGraph<T>::addGraphNode 的返回值.