c++ - Null 作为默认模板参数

Question

为什么不能在模板函数中使用 NULL 作为默认指针参数？ 让我们考虑以下代码:

template<class Graph, class NodeAttribs, class ArcAttribs> string
graphToGraphviz(Graph       &graph,
                NodeAttribs *nattribs = NULL,
                ArcAttribs  *aattribs = NULL,
                string      name      = ""){
   /*...*/
}

我希望能够这样调用它:

graphToGraphviz(g);

我怀疑编译器认为它无法解析 NULL 的类型，但如果属性为 NULL (有 if 条件)，则不会使用这些类型。但编译器可能无法以正确的方式解决这种情况。如果是，我怎样才能编写这样的重载函数，这将允许我使用简短的形式？

我有一个像这样重载它的想法:

class Empty{}

template<class Graph> string
graphToGraphViz(Graph       &graph,
                string      name      = ""){
    return graphToGraphviz<Graph, Empty, Empty>(graph, NULL, NULL, name)
}

但是编译器给了我错误，其中包括类 Empty 没有定义 operator [] 。这又不稳定，但是我是否必须制作所有这些“虚拟”运算符重载和空函数才能满足编译器的要求，或者是否有更好的方法来做到这一点？

编辑: 请查看完整的源代码 - 它将 Lemon graph 转换为 graphviz 格式: 我尝试使用 C++11 中的新语法(如下面的答案所示)，但没有成功。

#ifndef GRAPHTOGRAPHVIZ_H_
#define GRAPHTOGRAPHVIZ_H_

#include <lemon/list_graph.h>

using namespace lemon;
using namespace std;

/* USAGE:
 * ListDigraph::NodeMap<unordered_map<string, string>> nodeAttribs(g);
 * ListDigraph::ArcMap<unordered_map<string, string>> arcAttribs(g);
 * nodeAttribs[node]["label"] = "node_label";
 * string dot = graphToGraphviz(g, &nodeAttribs, &arcAttribs, "hello");
 */

template<class Map>
string getAttribs(Map &map){
    string attribs = "";
    for (const auto &el : map){
        if (el.second != "")
            attribs += "\"" + el.first + "\"=\"" + el.second + "\",";
    }
    if (attribs != "")
        attribs = " [" + attribs + "]";
    return attribs;
}


template<class Graph, class NodeAttribs, class ArcAttribs> string
graphToGraphviz(Graph       &graph,
                NodeAttribs *nattribs = NULL,
                ArcAttribs  *aattribs = NULL,
                string      name      = ""){

    typedef typename Graph::template NodeMap<string> NodeMap;
    typedef typename Graph::NodeIt NodeIterator;
    typedef typename Graph::ArcIt  ArcIterator;

    NodeMap labels(graph);
    ostringstream layout;
    layout << "strict digraph \""+name+"\" {\n";

    // prepare labels
    for (NodeIterator node(graph); node != INVALID; ++node){
        string label = "";
        if (*nattribs != NULL)
            label = (*nattribs)[node]["label"];
        if (label == "") label = static_cast<ostringstream*>( &(ostringstream() << graph.id(node)) )->str();
        label = "\"" + label + "\"";
        labels[node] = label;
    }

    // initialize nodes
    for (NodeIterator node(graph); node != INVALID; ++node){
        layout << labels[node];
        if (*nattribs != NULL)
            layout << getAttribs((*nattribs)[node]);
        layout << ";" << std::endl;
    }

    // initialize arcs
    for (ArcIterator arc(graph); arc != INVALID; ++arc){
        layout << labels[graph.source(arc)] << "->" << labels[graph.target(arc)];
        if (*aattribs != NULL)
            layout << getAttribs((*aattribs)[arc]);
        layout << ";" << std::endl;
    }
    layout << "}";
    return layout.str();
}


#endif /* GRAPHTOGRAPHVIZ_H_ */

使用 C++11 语法，函数头将如下所示:

template<class Graph, class NodeAttribs=ListDigraph::NodeMap<string>, class ArcAttribs=ListDigraph::NodeMap<string> > string
graphToGraphviz(Graph       &graph,
                NodeAttribs *nattribs = NULL,
                ArcAttribs  *aattribs = NULL,
                string      name      = "")

但它无法编译并给出大量奇怪的错误。

Answer 1

编译器在调用时出现问题:

graphToGraphviz(g);

现在 NodeAttribs 和 ArcAttribs 的类型是什么？
无论您是否使用它，编译器都必须推断出它的类型。因为使用或不使用是运行时检查。
使用您当前的代码，上述类型将变得不可推导。

how could I write such overloaded function

你的问题有答案了!!
重载模板函数，从原始模板函数中删除默认参数，并让两个函数共存:

template<class Graph>
string graphToGraphviz(Graph &graph, string name = "");