Question

我有以下功能对象：

/// 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一个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()));

我得到了

“错误：没有用于初始化的匹配构造函数 'const DOTf'（又名'const DOTWrapper'） const DOTf dotf（（NConv（）），（LConv（）））;“

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

使用未绑定的函数对象实例化函数对象会导致l值错误

0 个答案: