创建包装boost适配器的自定义范围类

时间:2018-10-17 21:41:52

标签: c++ boost c++17 c++20

我有一个类,将一些Boost转换适配器应用于范围(出于示例的原因,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,我们就可以使用以下范围来迭代Foo

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

但是,这与其他升压适配器或范围操作(例如boost::join)不能很好地结合在一起:

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

以上两种方法均会引起一些令人讨厌的模板错误。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不是范围。我尝试为建议here添加operator OutContainer() / iterator的{​​{1}}和typedef,但无济于事。我必须对const_iterator做些什么才能使其在这些范围操作中正常发挥作用?

1 个答案:

答案 0 :(得分:2)

在这种情况下,错误非常有用。您的类型必须建模SinglePassRange,而不必。有一个page关于如何执行此操作,虽然您提供了begin()end(),但没有为iteratorconst_iterator提供类型别名。因此,您不会建模SinglePassRange

但是您的代码失败实际上是非常好的,因为它也是不好的。您有begin()个呼叫range().begin()end()个呼叫range().end()。这些是不同范围中的迭代器。因此,无论如何,这都是未定义的行为-您无法满足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; });

这意味着您只需要编写一个函数:range()