5

这似乎是一件相当基本的事情,所以我正在寻找一个或多或少的简短、内置且易于阅读的解决方案。
我设法怀孕的最短的事情是

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> ) ) );
        } );

这远非简短易读......

4

1 回答 1

3

正如@jv_ 所指出的那样,hana::replicate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() { }
于 2016-10-18T15:25:27.357 回答