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我有以下函数对象：

/// function object wrapping node and label data into strings of DOT code
/// requires as template parameters 2 function objects that define behaviour
/// converting node and lable into strings. Requires as first template
/// argument the graph edge type
template<typename GRAPHEDGE, typename NodeToStringConverter,
         typename LabelToStringConverter>
struct DOTWrapper
{

    typedef NodeToStringConverter NodeConverter;
    typedef LabelToStringConverter LabelConverter;

    /// initialises to-string-converter objects
    DOTWrapper(NodeConverter& nc, LabelConverter& lc): nts(nc), lts(lc){}

    std::string start() const
    {
        return ("digraph  {\n");
    }

    // wrap an egde into dot code
    std::string wrap_edge(const GRAPHEDGE& e) const
    {
        // n needs to be to_string-able
        return ("\t\""+nts(e.source())+"\""+" -> "+"\""+nts(e.target())+"\""+
                " [ label = \""+lts(e.label())+"\" ];\n");
    }

    std::string finish() const { return "}\n"; }

    const NodeToStringConverter nts;
    const LabelToStringConverter lts;

};

template<typename Node>
struct NTSint
{

    std::string operator()(const Node& node) const { return node; }
};

template<typename Label>
struct LTSstr
{

    std::string operator()(const Label& label) const { return label; }
};

当我尝试像这样实例化 Wrapper 对象时：

typedef NTSint<Edge::Node> NConv;
typedef LTSstr<Edge::Label> LConv;
typedef DOTWrapper<Edge, NConv, LConv> DOTf;
const DOTf dotf((NConv()), (LConv()));

我明白了

“错误：没有匹配的构造函数用于初始化'const DOTf'（又名'const DOTWrapper'）const DOTf dotf（（NConv（）），（LConv（）））；”

这是为什么？NConv 和 LConv 毕竟在内存中占有一席之地......所以......它们不应该是左值吗？

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