Gcc C+中元组的递归打印+；

关于如何打印

元组

，有几种建议。下面的代码片段从重载运算符中复制了Kenny的答案您可以在此处了解clang不编译它的原因：。看来，克朗的行为符合标准

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为了解决这个问题，您可以定义

操作符，下面是一个使用integer_序列的示例：只想指出链接的答案并不能解决编译问题-它仍然失败。这只是打印元组的另一种技术。糟糕，问题当然是
名称空间
：（我怎么会忘记这么明显的事情。谢谢！
#include <iostream>
#include <tuple>
#include <type_traits>

// template <typename... T>
// std::ostream& operator<<(std::ostream& os, const std::tuple<T...>& tup);

template <size_t n, typename... T>
typename std::enable_if<(n >= sizeof...(T))>::type
print_tuple(std::ostream&, const std::tuple<T...>&)
{}

template <size_t n, typename... T>
typename std::enable_if<(n < sizeof...(T))>::type
print_tuple(std::ostream& os, const std::tuple<T...>& tup)
{
  if (n != 0)
    os << ", ";
  os << std::get<n>(tup);
  print_tuple<n+1>(os, tup);
}

template <typename... T>
std::ostream& operator<<(std::ostream& os, const std::tuple<T...>& tup)
{
  os << "[";
  print_tuple<0>(os, tup);
  return os << "]";
}

int
main()
{
  auto t = std::make_tuple(1, std::make_tuple(2, 3));
    std::cout << t << std::endl;
}

clang++-mp-3.5 -std=c++11 print.cc 
print.cc:19:10: error: call to function 'operator<<' that is neither visible in the
      template definition nor found by argument-dependent lookup
      os << std::get<n>(tup);
         ^
print.cc:20:7: note: in instantiation of function template specialization
      'print_tuple<1, int, std::__1::tuple<int, int> >' requested here
      print_tuple<n+1>(os, tup);
      ^
print.cc:27:7: note: in instantiation of function template specialization
      'print_tuple<0, int, std::__1::tuple<int, int> >' requested here
      print_tuple<0>(os, tup);
      ^
print.cc:35:19: note: in instantiation of function template specialization
      'operator<<<int, std::__1::tuple<int, int> >' requested here
        std::cout << t << std::endl;
                  ^
print.cc:24:19: note: 'operator<<' should be declared prior to the call site
    std::ostream& operator<<(std::ostream& os, const std::tuple<T...>& tup)
              ^