Question

我使用an answer to the SO question "iterate over tuple"编写了一个重载<<的方法。这个方法已经过测试，似乎可以在Debian squeeze上与g++ 4.7一起正常工作。

然而，这种方法有点迂回，因为似乎<<无法显式实例化（我发现了一篇关于它的帖子 here）。因此，一个人被迫定义一个字符串方法，然后调用它。我有一个类似的矢量方法，更直接。有没有人建议如何消除创建字符串方法的额外步骤，使用相同的方法，或其他？提前谢谢。

#include <tuple>
#include <iostream>
#include <string>
#include <sstream>
#include <vector>

using std::ostream;
using std::cout;
using std::endl;
using std::vector;
using std::string;

// Print vector<T>.
template<typename T> ostream& operator <<(ostream& out, const vector<T> & vec)
{
  unsigned int i;
  out << "[";
  for(i=0; i<vec.size(); i++)
    {
      out << vec[i];
      if(i < vec.size() - 1)
    out << ", ";
    }
  out << "]";
  return out;
}

////////////////////////////////////////////////////////////////

// Print tuple.
template<std::size_t I = 0, typename... Tp>
inline typename std::enable_if<I == sizeof...(Tp), string>::type
stringval(const std::tuple<Tp...> & t)
{
  std::stringstream buffer;
  buffer << "]";
  return buffer.str();
}

template<std::size_t I = 0, typename... Tp>
inline typename std::enable_if<I < sizeof...(Tp), string>::type
stringval(const std::tuple<Tp...> & t)
{
  std::stringstream buffer;
  size_t len = sizeof...(Tp);
  if(I==0)
      buffer << "[";
  buffer << std::get<I>(t);
  if(I < len - 1)
    buffer << ", ";
  buffer << stringval<I + 1, Tp...>(t);
  return buffer.str();
}

template<typename... Tp> ostream& operator <<(ostream& out, const std::tuple<Tp...> & t)
{
  out << stringval(t);
  return out;
}

int
main()
{
  typedef std::tuple<int, float, double> T;
  std::tuple<int, float, double> t = std::make_tuple(2, 3.14159F, 2345.678);
  cout << t << endl;
}

编译时，这会给出

[2, 3.14159, 2345.68]

Answer 1

您只需将std::ostream&传递到该stringval函数中，然后使用out <<代替buffer <<。

Demo：

#include <tuple>
#include <iostream>
#include <type_traits>

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 << "]";
}

Answer 2

基于折叠表达式（C ++ 17），索引序列（C ++ 14），lambda函数和模板参数包（均为C）的非递归 C ++ 17 -way解决方案++ 11）：

#include <tuple>
#include <iostream>
#include <ostream>
#include <utility>

template< typename F, typename ...types >
F
for_all(F f, types &&... values)
{
    (f(std::forward< types >(values)), ...);
    return std::move(f);
}

template< typename F, typename ...types, std::size_t ...indices >
F
for_all_indices(F f, std::tuple< types... > const & t, std::index_sequence< indices... >)
{
    return for_all(std::move(f), std::get< indices >(t)...);
}

template< typename first, typename ...rest > // non-nullary tuples only
std::ostream &
operator << (std::ostream & out, std::tuple< first, rest... > const & t)
{
    //return ((out << std::get< first >(t)) << ... << std::get< rest >(t)); // simply prints extracted tuple elements w/o delimiters
    out << '[';
    for_all_indices([&out] (auto const & value) { out << value << ", "; }, t, std::index_sequence_for< rest... >{});
    return out << std::get< sizeof...(rest) >(t) << ']';
}

int
main()
{
    std::cout << std::make_tuple(1, 2.F, 3.0) << std::endl;
    return 0;
}

LIVE DEMO

Answer 3

可能你不需要C ++ 17（尚未发布）来获得非递归（实际上是递归的，但以更自然的方式）解决方案。即，您不需要折叠表达式，只需要索引序列（C ++ 14）和模板参数包（C ++ 11）。

#include <iostream>
#include <sstream>
#include <utility>
#include <tuple>
#include <string>

template<class T>
std::ostringstream& concat_to_stream(std::ostringstream &oss, T &&arg) {
  oss << arg;
  return oss;
}

template<class First, class ...Rest>
std::ostringstream& concat_to_stream(std::ostringstream &oss, First &&firstArg, Rest &&... restArgs) {
  oss << firstArg << ", ";
  return concat_to_stream(oss, std::forward<Rest &&>(restArgs)...);
}

template<class ...Types>
std::string concat_to_string(Types &&... args) {
  std::ostringstream oss;
  oss << '[';
  concat_to_stream(oss, std::forward<Types &&>(args)...);
  oss << ']';
  return oss.str();
}

template<class Tuple, size_t... Indices>
std::string help_concat(const Tuple &tuple, std::index_sequence<Indices...>) {
  return concat_to_string(std::get<Indices>(tuple)...);
};

template<class ...Types>
std::string tuple_to_string(const std::tuple<Types...> &tuple) {
  return help_concat(tuple, std::make_index_sequence<sizeof...(Types)>{});
};

template<class ...Types>
std::ostream &operator<<(std::ostream &os, const std::tuple<Types...> &tuple) {
  return os << tuple_to_string(tuple);
}

int main() {
  std::tuple<int, double, std::string> sample_tuple = std::make_tuple(3, 1.723, "Hi!");
  std::cout << sample_tuple << '\n'; // [3, 1.723, Hi!]
  return 0;
}

递归部分是concat_to_stream部分，这很自然而且很常见。关键部分是help_concat，我从Implementing std::tuple From The Ground Up: Part 6, tuple_cat Take 1学习。

该技术是在参数列表中使用虚拟std::index_sequence来推导模板参数列表中的size_t... Indices，从而允许我们“平坦”＃34;将std::tuple的内容放入可变参数列表中，concat_to_string函数可以接受该列表。

Answer 4

这是一个非递归版本，使用std :: integer_sequence和其他一些相关技术。

template<class Ch, class Tr, class Tuple, std::size_t... Is>
void print_tuple_impl(std::basic_ostream<Ch,Tr>& os,
                      const Tuple& t,
                      std::index_sequence<Is...>)
{
    using swallow = int[];
    (void)swallow{0, (void(os << (Is == 0? "" : ", ") << std::get<Is>(t)), 0)...};
}

template<class Ch, class Tr, class... Args>
decltype(auto) operator<<(std::basic_ostream<Ch, Tr>& os,
                          const std::tuple<Args...>& t)
{
    os << "(";
    print_tuple_impl(os, t, std::index_sequence_for<Args...>{});
    return os << ")";
}

原来它来自这里： http://en.cppreference.com/w/cpp/utility/integer_sequence

重载运算符＆lt;＆lt;对于std :: tuple - 可能的简化？

4 个答案: