如何为Hana序列写一个for循环？

本文介绍了如何为Hana序列写一个for循环？的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我有一个Boos.Hana序列，我想打印它到屏幕，用逗号分隔。但是逗号只分开元素，所以我必须检查我是否在最后一个元素。

目前我的黑客是非常糟糕的（看着指针，并转换到 void * 。

template< class P，class ... Ts> decltype（auto）operator<<（ std :: ostream& os， boost :: hana :: tuple< Ts ...> const& tpl ）{ os<<{; boost： ：hana :: for_each（ tpl，[& x] { os<< x; if（（void *）&boost; hana： ：back（tpl）！=（void *）& x）os<<，; } ）; return os< b $ b}

在Boost.Fusion的情况下，它更复杂，因为我使用融合迭代器 boost :: fusion :: begin 和 boost :: fusion :: end ），但至少我可以比较迭代器。（ bool last = result_of :: equal_to< typename result_of :: next< First> :: type，Last> :: value ）。

另一种提出这个问题的方法是如果在Hana中有（meta）迭代器。

解决方案

，为了回答你的意见， drop_back 确实做了一个副本。 Hana中的所有算法都制作副本，并且渴望，如此处所述。 p>

其次，您可以使用 hana :: intersperse 在每个元素之间添加逗号，导致类似

template< class P，class ... Ts> decltype（auto）operator<<（ std :: ostream& os， boost :: hana :: tuple< Ts ...> const& tpl ） { os<< {; boost :: hana :: for_each（boost :: hana :: intersperse（tpl，，）， [&]（auto const& x）{ os< ; x; }）; return os<< };然而，最好的解决方案可能是使用 }

< experimental :: print ，这完全符合您的需求：

#include< boost / hana /experimental/printable.hpp> #include< boost / hana / tuple.hpp> #include< iostream> int main（）{ auto ts = hana :: make_tuple（1，2，3）; std :: cout<< hana :: experimental :: print（ts）; }

/ strong>

如果要使用 intersperse 解决方案，但不想复制可以执行以下操作：

#include< boost / hana.hpp> #include< functional> #include< iostream> namespace hana = boost :: hana; template< class ... Ts> decltype（auto）operator<<（std :: ostream& os，hana :: tuple< Ts ...> const& tpl）{ os& {; char const * sep =，; auto refs = hana :: transform（tpl，[]（auto const& t）{return std :: ref（t）;}）; hana :: for_each（hana :: intersperse（refs，std :: ref（sep））， [&]（auto const& x）{ os<< x .get（）; }）; return os<< }; }

但实际上你应该使用 hana： ：experimental :: print 。如果你的用例是性能关键的，你想避免创建一个 std :: string ，我会质疑 std :: ostream 。

编辑结束

I have a Boos.Hana sequence and I would like to print it to screen separated by commas. However the commas separate elements only, so I have to check if I am at the last element.

Currently my hack is pretty bad (looking at the pointer and casting to void*.

template<class P, class... Ts> decltype(auto) operator<<( std::ostream& os, boost::hana::tuple<Ts...> const& tpl ){ os << "{"; boost::hana::for_each( tpl, [&](auto& x){ os << x; if((void*)&boost::hana::back(tpl) != (void*)&x) os << ", "; } ); return os << "}"; }

In the case of Boost.Fusion it was more complicated because I use fusion iterators (boost::fusion::begin and boost::fusion::end) but at least I could compare the iterators. (bool last = result_of::equal_to<typename result_of::next<First>::type, Last>::value).

Another way to ask this question is if there are (meta) iterators in Hana.

解决方案

First, to answer your comment, drop_back does make a copy. All algorithms in Hana make copies and are eager, as documented here.

Secondly, you could use hana::intersperse to add a comma between each element, resulting in something like

template<class P, class... Ts> decltype(auto) operator<<( std::ostream& os, boost::hana::tuple<Ts...> const& tpl ){ os << "{"; boost::hana::for_each(boost::hana::intersperse(tpl, ", "), [&](auto const& x){ os << x; }); return os << "}"; }

However, the best solution would probably be to use experimental::print, which does exactly what you want:

#include <boost/hana/experimental/printable.hpp> #include <boost/hana/tuple.hpp> #include <iostream> int main() { auto ts = hana::make_tuple(1, 2, 3); std::cout << hana::experimental::print(ts); }

Edit

If you want to use the intersperse solution, but do not want to make a copy of the sequence, you can do the following:

#include <boost/hana.hpp> #include <functional> #include <iostream> namespace hana = boost::hana; template <class... Ts> decltype(auto) operator<<(std::ostream& os, hana::tuple<Ts...> const& tpl) { os << "{"; char const* sep = ", "; auto refs = hana::transform(tpl, [](auto const& t) { return std::ref(t); }); hana::for_each(hana::intersperse(refs, std::ref(sep)), [&](auto const& x){ os << x.get(); }); return os << "}"; }

But really, you should probably be using hana::experimental::print. And if your use case is performance critical and you want to avoid creating a std::string, I would question the usage of std::ostream in the first place.

End of edit