c++ - 如何确定迭代器在 C++ 模板函数中指向的对象的类型？

Question

我在 C++ 中有一个模板函数，它序列化一个可迭代对象:

template<typename Stream, typename Iter, typename Infix, typename Closure>
inline Stream &stream_iterable(Stream &os, Iter from, Iter to, Infix infix_, Closure open, Closure close) {
    if (from == to) return os;
    os << open << *from;
    for (++from; from != to; ++from) {
        os << infix_ << *from;
    }
    os << close;
    return os;
}

例如，它基本上转换std::vector<int>{1,2}到一个字符串 "[1,2]"

我想检查迭代器指向的对象类型，如果它是 std::string , 我想用 std::quoted在 vector 的元素周围添加引号，像这样:

template<typename Stream, typename Iter, typename Infix, typename Closure>
inline Stream &steam_iterable_quoted(Stream &os, Iter from, Iter to, Infix infix_, Closure open, Closure close) {
    if (from == to) return os;
    os << open << std::quoted(*from);
    for (++from; from != to; ++from) {
        os << infix_ << std::quoted(*from);
    }
    os << close;
    return os;
}

如何检查 (*from) 的类型并将这两个函数合二为一？

Answer 1

您实际上不需要知道 stream_iterable 主体中的类型。俗话说，加一层间接:

namespace detail {
  template<typename T>
  constexpr T const& prepare_to_print(T const& t) { return t; }

  auto prepare_to_print(std::string const s&) { return std::quoted(s); }
  // A non-const overload would be required as well...
  // Forwarding can be a drag sometimes 
}

只需将取消引用的迭代器传递给 prepare_to_print。重载的好处是您可以通过稍后添加更多重载来进一步自定义行为。