将迭代器取消引用到临时范围时出现非指针操作数错误 [英] Non-pointer-operand error when dereferencing an iterator into a temporary range

问题描述

使用

auto empty_line = [](auto& str){ return str.size() == 0; };

我们可以这样做:

auto line_range_with_first_non_empty = 
                ranges::view::drop_while(ranges::getlines(std::cin),empty_line);
auto input1 = std::stoi(*line_range_with_first_non_empty.begin());

我们也可以这样做:

auto line_range2 = ranges::getlines(std::cin);
auto iter2 = ranges::find_if_not(line_range2,empty_line);
auto input2 = std::stoi(*iter2);

不幸的是，当我尝试将以上版本缩短为:

Unfortunately, when I try to shorten version above into:

auto iter3 = ranges::find_if_not(ranges::getlines(std::cin),empty_line);
// auto input3 = std::stoi(*iter3);

我得到一个错误:

<source>:22:29: error: indirection requires pointer operand ('ranges::v3::dangling<ranges::v3::_basic_iterator_::basic_iterator<ranges::v3::getlines_range::cursor> >' invalid)
    auto input3 = std::stoi(*iter3);
                            ^~~~~~

我认为是因为存在无限范围，但我错了.

I thought it's because of that infinite range but I was wrong.

auto sin = std::istringstream{"\n\n\nmy line\n"};
auto iter4 = ranges::find_if_not(ranges::getlines(sin),empty_line);
// Error when deref.
// auto input4 = std::stoi(*iter4);

这会产生相同的错误.

<source>:27:29: error: indirection requires pointer operand ('ranges::v3::dangling<ranges::v3::_basic_iterator_::basic_iterator<ranges::v3::getlines_range::cursor> >' invalid)
    auto input4 = std::stoi(*iter4);
                        ^~~~~~

当ranges::find_if将范围作为右值时，为什么不能取消引用?

Why can't I dereference when ranges::find_if takes a range as rvalue?

ranges::getlines是否返回范围?如果是这样，范围应该拥有东西吗?

Does ranges::getlines return a range? If so, are ranges supposed to own things?

godbolt.org/g/Yo6tKa

推荐答案

如果传递给算法的范围是临时的，并且该算法返回迭代器，则将该迭代器包装在dangling包装器中，以防止您执行此操作任何不安全的东西.任务完成. :-)

If the range passed to an algorithm is a temporary, and the algorithm returns an iterator, the iterator is wrapped in a dangling wrapper to keep you from doing anything unsafe. Mission accomplished. :-)

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