view :: join是否需要可复制的内部范围？为什么？

Question

假设我们有

cppcoro::generator<int> gen_impl(int in) {
  const auto upper = in + 10;
  for (; in < upper; ++in)
    co_yield in;
}

cppcoro::generator<cppcoro::generator<int>> gen() {
  for (int n = 1; n < 100; n += 10)
    co_yield gen_impl(n);
}

所以我们可以很好地迭代内部范围

  for (auto&& row : gen() ) {
    for (auto n : row)
      std::cout << n << ' ';
    std::cout << '\n';
  }

注意:范围是for ref是必需的,因为cppcoro::generator不允许复制(删除的副本ctor)

打印

1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16 17 18 19 20
21 22 23 24 25 26 27 28 29 30
31 32 33 34 35 36 37 38 39 40
41 42 43 44 45 46 47 48 49 50
51 52 53 54 55 56 57 58 59 60
61 62 63 64 65 66 67 68 69 70
71 72 73 74 75 76 77 78 79 80
81 82 83 84 85 86 87 88 89 90
91 92 93 94 95 96 97 98 99 100

但是当我们尝试使用view :: join"flattern"时

auto rng = gen();
for (auto n : rng | ranges::view::join) {
  std::cout << n << '\n';
};

似乎view :: join需要可复制的内部范围？

In file included from <source>:3:

In file included from /opt/compiler-explorer/libs/rangesv3/trunk/include/range/v3/view.hpp:38:

In file included from /opt/compiler-explorer/libs/rangesv3/trunk/include/range/v3/view/for_each.hpp:23:

/opt/compiler-explorer/libs/rangesv3/trunk/include/range/v3/view/join.hpp:320:50: error: call to deleted constructor of 'cppcoro::generator<cppcoro::generator<int> >'

                    return join_view<all_t<Rng>>{all(static_cast<Rng&&>(rng))};

                                                 ^~~~~~~~~~~~~~~~~~~~~~~~~~~~

/opt/compiler-explorer/libs/rangesv3/trunk/include/range/v3/view/view.hpp:112:21: note: in instantiation of function template specialization 'ranges::v3::view::join_fn::operator()<cppcoro::generator<cppcoro::generator<int> > &, false, nullptr>' requested here

                    v.view_(static_cast<Rng&&>(rng))

                    ^

/opt/compiler-explorer/libs/rangesv3/trunk/include/range/v3/utility/functional.hpp:731:42: note: in instantiation of function template specialization 'ranges::v3::view::view<ranges::v3::view::join_fn>::pipe<cppcoro::generator<cppcoro::generator<int> > &, ranges::v3::view::view<ranges::v3::view::join_fn> &, false, nullptr>' requested here

            pipeable_access::impl<Pipe>::pipe(static_cast<Arg&&>(arg), pipe)

                                         ^

<source>:35:21: note: in instantiation of function template specialization 'ranges::v3::operator|<cppcoro::generator<cppcoro::generator<int> > &, ranges::v3::view::view<ranges::v3::view::join_fn>, false, nullptr>' requested here

  for (auto n : rng | ranges::view::join) {

                    ^

/opt/compiler-explorer/libs/cppcoro/include/cppcoro/generator.hpp:174:3: note: 'generator' has been explicitly marked deleted here

                generator(const generator& other) = delete;

                ^

/opt/compiler-explorer/libs/rangesv3/trunk/include/range/v3/view/join.hpp:76:36: note: passing argument to parameter 'rng' here

            explicit join_view(Rng rng)

                                   ^

是什么让这不编译？

range-v3或cppcoro中是否有任何错误？

只有不兼容的设计决策？

godbolt(全)

Answer 1

在range-v3中,仅移动视图正常.这实施得很晚,可能仍然存在错误,但这不是这里发生的事情.

第一个问题是你试图在cppcoro::generator这里调整左值类型:

auto rng = gen();
for (auto n : rng | ranges::view::join) {

由于生成器是视图,因此join视图将要复制它.它不能,因为它不可复制.

您可以通过移动生成器来解决此问题:

auto rng = gen();
for (auto n : std::move(rng) | ranges::view::join) {

然后你遇到下一个问题,就是引用类型generator<generator<int>>是const generator<int>&,你再次遇到同样的问题:join想要在迭代它时保留内部生成器的副本,但它不能复制.

解决方法有点难看:更改生成器以返回非const左值引用:

cppcoro::generator<cppcoro::generator<int>&> gen() {
  for (int n = 1; n < 100; n += 10) {
    auto tmp = gen_impl(n);
    co_yield tmp;
  }
}

然后std::move每个内部范围都有一个move视图:

auto rng = gen();
for (auto n : std::move(rng) | ranges::view::move | ranges::view::join) {
  std::cout << n << '\n';
}

结果编译.无论它是否运行取决于cppcoro如何优雅地处理有人偷走了它在coroutine的承诺类型中安全隐藏的价值的内容.

https://godbolt.org/z/mszidX

关于未来的说明std::view::join:

join随C++ 20提供的视图略有不同.如果外部范围的引用类型是实际引用(如本例所示),则它不会尝试复制它引用的视图.这意味着在C++ 20中,你不需要丑陋的view::move黑客攻击.

但是,C++ 20 View概念目前需要可复制性,因此该解决方案仍然不起作用.我们有一个TODO项目可以在C++ 20发布之前放松一下,但是没有人知道委员会会如何看待这个想法.