c++ - 打印非类型模板内部类

Question

我有一个非类型类模板(例如 Counted)，它指向类模板的一个实例(例如 Counter)。只要两个模板类都是兄弟，一切都可以正常工作，例如编写一个打印操作符。

现在由于几个原因，我想将 Counted 作为内部类移到 Counter 内部，但我发现自己无法编写打印运算符。

Here's a working "sibling class" version , 主要代码在这里:

template < class Count >
struct Counter {
    using count_type = Count;
    count_type instances = 0;
};

template < class Cnt, Cnt& c>
struct Counted {
    using id_type = typename Cnt::count_type;
    id_type id;
    Counted() : id(c.instances) { ++c.instances; }
    ~Counted() { --c.instances; }
};

template < class Cnt, Cnt& c >
std::ostream& operator<<(std::ostream& os, Counted<Cnt,c> sib) {
    os << "printing sibling " << sib.id;
    return os;
}

using SCounter = Counter<std::size_t>;
SCounter sc;
using SCounted = Counted<SCounter, sc>;

Here's the not compiling "inner class" version主要代码在这里:

template < class Count >
struct Counter {
    using count_type = Count;
    count_type instances = 0;

    template <Counter& c>
    struct Counted {
        using id_type = count_type;
        id_type id;
        Counted() : id(c.instances) { ++c.instances; }
        ~Counted() { --c.instances; }
    };
};

template < class Count, Counter<Count>& c >
std::ostream& operator<<(std::ostream& os,
      typename Counter<Count>::template Counted<c>& sib) {
    os << "printing inner " << sib.id;
    return os;
}

using SCounter = Counter<std::size_t>;
SCounter sc;
using SCounted = SCounter::Counted<sc>;

错误:

prog.cpp: In function 'int main()':
prog.cpp:32:15: error: cannot bind 'std::ostream {aka std::basic_ostream<char>}' lvalue to 'std::basic_ostream<char>&&'
  std::cout << j << std::endl;
               ^
In file included from /usr/include/c++/4.9/iostream:39:0,
                 from prog.cpp:1:
/usr/include/c++/4.9/ostream:602:5: note: initializing argument 1 of 'std::basic_ostream<_CharT, _Traits>& std::operator<<(std::basic_ostream<_CharT, _Traits>&&, const _Tp&) [with _CharT = char; _Traits = std::char_traits<char>; _Tp = Counter<unsigned int>::Counted<(* & sc)>]'
     operator<<(basic_ostream<_CharT, _Traits>&& __os, const _Tp& __x)
     ^

我的代码、我的编译器是否有问题，或者这是标准不允许的？

Answer 1

Koenig 运算符就是您想要的:

  template <Counter& c>
    struct Counted {
      using id_type = count_type;
      id_type id;
      Counted() : id(c.instances) { ++c.instances; }
      ~Counted() { --c.instances; }

      friend std::ostream& operator<<(
        std::ostream& os,
        Counted const& sib
      ) {
        os << "printing inner " << sib.id;
        return os;
      }           
    };
};

这里我们创建一个 friend operator<< .当您尝试使用 << 时，它将通过 ADL(或 Koenig 查找)找到用Counted实例。

为每种类型的 Counted 创建了一个不同的函数，并且没有进行模板推导。 .