std::ranges::prev
<iterator>
constexpr I prev( I i );
constexpr I prev( I i, std::iter_difference_t <I> n );
constexpr I prev( I i, std::iter_difference_t <I> n, I bound );
Return the nth predecessor of iterator i.
The function-like entities described on this page are algorithm function objects (informally known as niebloids), that is:
- Explicit template argument lists cannot be specified when calling any of them.
- None of them are visible to argument-dependent lookup.
- When any of them are found by normal unqualified lookup as the name to the left of the function-call operator, argument-dependent lookup is inhibited.
Contents
[edit] Parameters
[edit] Return value
[edit] Complexity
[edit] Possible implementation
struct prev_fn { template<std::bidirectional_iterator I> constexpr I operator()(I i) const { --i; return i; } template<std::bidirectional_iterator I> constexpr I operator()(I i, std::iter_difference_t <I> n) const { ranges::advance (i, -n); return i; } template<std::bidirectional_iterator I> constexpr I operator()(I i, std::iter_difference_t <I> n, I bound) const { ranges::advance (i, -n, bound); return i; } }; inline constexpr auto prev = prev_fn();
[edit] Notes
Although the expression --r.end() often compiles for containers, it is not guaranteed to do so: r.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, --r.end() does not compile, while ranges::prev(r.end()) does.
This is further exacerbated by ranges that do not model ranges::common_range . For example, for some underlying ranges, ranges::transform_view::end doesn't have the same return type as ranges::transform_view::begin, and so --r.end() won't compile. This isn't something that ranges::prev can aid with, but there are workarounds.
[edit] Example
#include <iostream> #include <iterator> #include <vector> int main() { std::vector <int> v{3, 1, 4}; auto pv = std::ranges::prev(v.end(), 2); std::cout << *pv << '\n'; pv = std::ranges::prev(pv, 42, v.begin()); std::cout << *pv << '\n'; }
Output:
1 3