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Non-pointer-operand error when dereferencing an iterator into a temporary range

Using

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

we can do this:

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());

and we can also do this:

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

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);

I get an error:

<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);
                            ^~~~~~

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);

This produces the same error.

<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);
                        ^~~~~~

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

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

godbolt.org/g/Yo6tKa

like image 416
sandthorn Avatar asked Jan 28 '23 14:01

sandthorn


1 Answers

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. :-)

like image 179
Eric Niebler Avatar answered Feb 06 '23 15:02

Eric Niebler