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Showing content from https://en.cppreference.com/w/cpp/language/../numeric/../string/../iterator/prev.html below:

std::prev - cppreference.com

Return the n^th predecessor (or -n^th successor if n is negative) of iterator it.

An iterator of type BidirIt that holds the n^th predecessor (or -n^th successor if n is negative) of iterator it.

Linear.

However, if BidirIt additionally meets the requirements of LegacyRandomAccessIterator, complexity is constant.

template<class BidirIt>
constexpr // since C++17
BidirIt prev(BidirIt it, typename std::iterator_traits<BidirIt>::difference_type n = 1)
{
    std::advance(it, -n);
    return it;
}

Although the expression --c.end() often compiles, it is not guaranteed to do so: c.end() is an rvalue expression, and there is no iterator requirement that specifies that decrement of an rvalue is guaranteed to work. In particular, when iterators are implemented as pointers or its operator-- is lvalue-ref-qualified, --c.end() does not compile, while std::prev(c.end()) does.

#include <iostream>
#include <iterator>
#include <vector>
 
int main()
{
    std::vector<int> v{3, 1, 4};
 
    auto it = v.end();
    auto pv = std::prev(it, 2);
    std::cout << *pv << '\n';
 
    it = v.begin();
    pv = std::prev(it, -2);
    std::cout << *pv << '\n';
}

Output:

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