Question

这似乎是一个相当基本的事情，所以我正在寻找一个或多或少的简短，内置和易读的解决方案。
我设想的最短的事情是

hana::unfold_left<hana::tuple_tag>( hana::int_c<n>, [] ( auto count ) {
            return hana::if_( count == hana::int_c<0>, hana::nothing,
                             hana::just( hana::make_pair( count - hana::int_c<1>,
                                                        hana::type_c<T> ) ) );
        } );

远非简短易读......

Answer 1

正如@jv_ hana::replicate所指出的，可以做到这一点参考文档中的示例提供了有关如何实现此目的的足够信息：

// Copyright Louis Dionne 2013-2016
// Distributed under the Boost Software License, Version 1.0.
// (See accompanying file LICENSE.md or copy at http://boost.org/LICENSE_1_0.txt)

#include <boost/hana/equal.hpp>
#include <boost/hana/integral_constant.hpp>
#include <boost/hana/optional.hpp>
#include <boost/hana/replicate.hpp>
#include <boost/hana/tuple.hpp>

namespace hana = boost::hana;
static_assert(hana::replicate<hana::tuple_tag>('x', hana::size_c<2>) == hana::make_tuple('x', 'x'), "");
// Of course, there can't be more than one element in an `optional`.
static_assert(hana::replicate<hana::optional_tag>('x', hana::size_c<2>) == hana::just('x'), "");
int main() { }

如何建立一个hana :: tuple_t <t，t，=“”t，=“”... =“”>给定T和元素数量n

1 个答案: