创建一个包含升压适配器的自定义范围 class

Question

我有一个 class 将一些升压变换适配器应用到一个范围（例如，实际上它比这复杂得多）：

struct Foo {
    auto range() const {
        return boost::irange(0, 10)
            | boost::adaptors::transformed([] (auto x) { return x * 2; });
    }

    auto begin() const { return range().begin(); }
    auto end() const { return range().end(); }
};

仅此一项就允许我们使用以下范围迭代 Foo：

for (auto x : Foo()) {
    std::cout << num << std::endl;
}

但是，这不能很好地与其他增强适配器或范围操作（如 boost::join）组合：

auto bad = boost::join(boost::irange(0, 10), Foo());
auto also_bad = Foo() | boost::adaptors::transformed([] (auto x) { return x + 1; });

以上两者都会引发一些讨厌的模板错误。前者(bad):

In file included from test.cpp:4:
/usr/local/include/boost/range/join.hpp:30:70: error: no type named 'type' in
      'boost::range_iterator<const Foo, void>'
            BOOST_DEDUCED_TYPENAME range_iterator<SinglePassRange1>::type,
            ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^~~~
/usr/local/include/boost/range/join.hpp:44:28: note: in instantiation of template class
      'boost::range_detail::joined_type<const Foo, const boost::integer_range<int> >' requested
      here
    : public range_detail::joined_type<SinglePassRange1, SinglePassRange2>::type
                           ^
test.cpp:34:16: note: in instantiation of template class 'boost::range::joined_range<const Foo,
      const boost::integer_range<int> >' requested here
    auto bad = boost::join(Foo(), range);
               ^

...

而后者(also_bad):

In file included from test.cpp:1:
In file included from /usr/local/include/boost/range/any_range.hpp:17:
In file included from /usr/local/include/boost/range/detail/any_iterator.hpp:22:
In file included from /usr/local/include/boost/range/detail/any_iterator_wrapper.hpp:16:
In file included from /usr/local/include/boost/range/concepts.hpp:24:
/usr/local/include/boost/range/value_type.hpp:26:70: error: no type named 'type' in
      'boost::range_iterator<Foo, void>'
    struct range_value : iterator_value< typename range_iterator<T>::type >
                                         ~~~~~~~~~~~~~~~~~~~~~~~~~~~~^~~~
/usr/local/include/boost/range/adaptor/replaced.hpp:109:40: note: in instantiation of template
      class 'boost::range_value<Foo>' requested here
                BOOST_DEDUCED_TYPENAME range_value<SinglePassRange>::type>& f )
                                       ^
test.cpp:35:27: note: while substituting deduced template arguments into function template
      'operator|' [with SinglePassRange = Foo]
    auto also_bad = Foo() | boost::adaptors::transformed([] (auto x) { return x * 2; });
                          ^
...

这两个错误似乎都在抱怨 Foo 不是一个范围。我尝试按照建议添加 operator OutContainer() 和 iterator/const_iterator 的类型定义，但无济于事。我必须对 Foo 做些什么才能让它很好地处理这些范围操作？

Answer 1

这个错误在这种情况下非常有用。你的类型必须模仿 SinglePassRange and it does not. There's a page 如何做到这一点，虽然你提供了 begin() 和 end()，但你没有为 iterator 和 const_iterator 提供类型别名.因此，您不建模 SinglePassRange.

但是你的代码失败其实很好，因为它也很糟糕。您有 begin() 调用 range().begin() 和 end() 调用 range().end()。这些是 不同范围 的迭代器。所以这无论如何都是未定义的行为——你不符合 SinglePassRange.

的语义概念

更简单的解决方案是直接使用 Foo()。这已经有效：

auto good = boost::join(boost::irange(0, 10), Foo().range());
auto also_good = Foo().range() | boost::adaptors::transformed([] (auto x) { return x + 1; });

这意味着您只需编写一个函数：range()

创建一个包含升压适配器的自定义范围 class

Creating a custom range class that wraps boost adaptors

c++

boost

c++17

c++20