reverse_iterator assignment Expression:cannot seek vector iterator after end

A practice question on C++ primer Edition 5th:

> Given a vector that has ten elements,copy the elements from
> positions 3 through 7 in reverse order to a list.

#include <vector>
#include <iterator>
#include <iostream>
#include <list>
#include <algorithm>

using namespace std;

int main()
{
	list<int> ilst;
	vector<int> ivec{ 1,1,2,3,5,8,13,21,34,55 };
	auto iter = ivec.crbegin() - 2;
	
	for (; iter != ivec.crend() + 3; iter--)
	{
		auto it = back_inserter(ilst);
		*it = *iter;
	}
	return 0;
}

An error is reported when the program executes to assign a value to iter:
> Expression:cannot seek vector iterator after end.

What's wrong with the code?Can someone tell me where the problem is?

Reverse iterators go from end of the container to the beginning of the container. So calling ++ operator on the reverse iterator will move one position to left in the container.

#include <vector>
#include <iterator>
#include <iostream>
#include <list>
#include <algorithm>

using namespace std;

int main()
{
    list<int> ilst;
    vector<int> ivec{ 1,1,2,3,5,8,13,21,34,55 };
    for (auto it = ivec.crbegin() + 2; it != ivec.crbegin() + 7; it++)
    {
        ilst.push_back(*it);
    }

    for (auto x : ilst) cout << x << ' ';
    return 0;
}

Reverse iterators iterate containers in reverse order, it means that when moving a reverse iterator forward, it actually moves closer to the beginning of the container. Likewise, if you iterate it backwards, you get closer to the end of the container. It means that this line:

auto iter = ivec.crbegin() - 2;

Goes beyond the bounds of the container (2 elements past the last element in the container). Also this part:

ivec.crend() + 3

Goes 3 elements past the beginning of the container.

In both scenarios you refer to memory which the container doesn't own and it causes undefined behavior. What you apparently want instead is to take a reverse iterator referring to the 7'th element of the container:

auto it = std::prev(std::crend(source), 7)

And check that it never goes beyond a reverse iterator referring to the second element (in order to include the third one):

it != std::prev(std::crend(source), 2)

Here is how I would rewrite you solution:

#include <vector>
#include <iterator>
#include <list>
#include <algorithm>

int main() {
    std::list<int> target;
    const std::vector<int> source{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };

    for (auto it = source.size() > 7 ? std::prev(std::crend(source), 7) : std::crbegin(source);
         it != std::prev(std::crend(source), 2); ++it) {
        target.push_back(*it);
    }

    return 0;
}