漂亮的打印 std::tuple
这是我之前关于漂亮打印STL容器，为此我们设法开发了一个非常优雅且完全通用的解决方案.
This is a follow-up to my previous question on pretty-printing STL containers, for which we managed to develop a very elegant and fully general solution.
在下一步中，我想为 std::tuple<Args...> 包括使用可变参数模板的漂亮打印(所以这严格来说是 C++11).对于std::pair<S,T>，我简单说
In this next step, I would like to include pretty-printing for std::tuple<Args...>, using variadic templates (so this is strictly C++11). For std::pair<S,T>, I simply say
std::ostream & operator<<(std::ostream & o, const std::pair<S,T> & p)
{
return o << "(" << p.first << ", " << p.second << ")";
}
打印元组的类似结构是什么?
What is the analogous construction for printing a tuple?
我尝试了各种模板参数堆栈解包、传递索引并使用 SFINAE 来发现我何时位于最后一个元素，但没有成功.我不会用我损坏的代码给你带来负担；希望问题描述足够直截了当.本质上，我想要以下行为:
I've tried various bits of template argument stack unpacking, passing indices around and using SFINAE to discover when I'm at the last element, but with no success. I shan't burden you with my broken code; the problem description is hopefully straight-forward enough. Essentially, I'd like the following behaviour:
auto a = std::make_tuple(5, "Hello", -0.1);
std::cout << a << std::endl; // prints: (5, "Hello", -0.1)
包含与上一个问题相同级别的通用性(char/wchar_t，对分隔符)的加分！
Bonus points for including the same level of generality (char/wchar_t, pair delimiters) as the the previous question!
推荐答案
是的，indices~
namespace aux{
template<std::size_t...> struct seq{};
template<std::size_t N, std::size_t... Is>
struct gen_seq : gen_seq<N-1, N-1, Is...>{};
template<std::size_t... Is>
struct gen_seq<0, Is...> : seq<Is...>{};
template<class Ch, class Tr, class Tuple, std::size_t... Is>
void print_tuple(std::basic_ostream<Ch,Tr>& os, Tuple const& t, seq<Is...>){
using swallow = int[];
(void)swallow{0, (void(os << (Is == 0? "" : ", ") << std::get<Is>(t)), 0)...};
}
} // aux::
template<class Ch, class Tr, class... Args>
auto operator<<(std::basic_ostream<Ch, Tr>& os, std::tuple<Args...> const& t)
-> std::basic_ostream<Ch, Tr>&
{
os << "(";
aux::print_tuple(os, t, aux::gen_seq<sizeof...(Args)>());
return os << ")";
}
Ideone 上的实时示例.
对于分隔符，只需添加这些部分特化:
For the delimiter stuff, just add these partial specializations:
// Delimiters for tuple
template<class... Args>
struct delimiters<std::tuple<Args...>, char> {
static const delimiters_values<char> values;
};
template<class... Args>
const delimiters_values<char> delimiters<std::tuple<Args...>, char>::values = { "(", ", ", ")" };
template<class... Args>
struct delimiters<std::tuple<Args...>, wchar_t> {
static const delimiters_values<wchar_t> values;
};
template<class... Args>
const delimiters_values<wchar_t> delimiters<std::tuple<Args...>, wchar_t>::values = { L"(", L", ", L")" };
并相应地更改 operator<< 和 print_tuple:
and change the operator<< and print_tuple accordingly:
template<class Ch, class Tr, class... Args>
auto operator<<(std::basic_ostream<Ch, Tr>& os, std::tuple<Args...> const& t)
-> std::basic_ostream<Ch, Tr>&
{
typedef std::tuple<Args...> tuple_t;
if(delimiters<tuple_t, Ch>::values.prefix != 0)
os << delimiters<tuple_t,char>::values.prefix;
print_tuple(os, t, aux::gen_seq<sizeof...(Args)>());
if(delimiters<tuple_t, Ch>::values.postfix != 0)
os << delimiters<tuple_t,char>::values.postfix;
return os;
}
和
template<class Ch, class Tr, class Tuple, std::size_t... Is>
void print_tuple(std::basic_ostream<Ch, Tr>& os, Tuple const& t, seq<Is...>){
using swallow = int[];
char const* delim = delimiters<Tuple, Ch>::values.delimiter;
if(!delim) delim = "";
(void)swallow{0, (void(os << (Is == 0? "" : delim) << std::get<Is>(t)), 0)...};
}
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