好吧,我想知道为什么endd似乎没有执行(尽管它在编译时不会产生任何错误). struct dxfDato{ dxfDato(int c, string v = 0, int t = 0) { codigo = c; valor = v; tipo = t; } dxfDato() { } int tipo; int codigo; string valor
struct dxfDato { dxfDato(int c, string v = 0, int t = 0) { codigo = c; valor = v; tipo = t; } dxfDato() { } int tipo; int codigo; string valor; }; class dxfItem { private: std::ostringstream ss; typedef std::ostream& (*manip)(std::ostream&); public: int clase; string valor; vector<dxfDato> datos; vector<dxfItem> hijos; template <typename T> dxfItem& operator<<(const T& x) { ss << x; return *this; } dxfItem& operator<<(manip x) // to store std manipulators { ss << x; return *this; } static dxfItem& endd(dxfItem& i) // specific manipulator 'endd' { dxfDato dd; dd.valor = i.ss.str(); i.datos.push_back(dd); std::cout << "endd found!" << endl; return i; } }; /* blah blah blah */ dxfItem header; header << 9 << endl << "$ACADVER" << endl << 1 << endl << "AC1500" << endl << dxfItem::endd // this apparently doesn't execute anything << "other data" << endl ;
这是我在尝试开发某些东西时发现的最后一个问题.最后一件事暴露在这里:C++ Operator overloading example
谢谢你们!
您已经将类型操作定义为通过引用获取std :: ostream并通过引用返回std :: ostream的函数,但是您已定义endd以获取dxfItem并返回dxfItem并且dxfItem不是从std派生的:: ostream的.由于这种类型不匹配,编译器正在生成对运算符的调用<<模板,而不是操纵过载.
此外,你的操作重载需要实际调用传递给它的操纵器函数:
dxfItem& operator<<(manip x) { x(ss); return *this; }