std::prev

From cppreference.com
< cpp‎ | iterator
Defined in header <iterator>
template< class BidirIt >

BidirIt prev(
  BidirIt it,

  typename std::iterator_traits<BidirIt>::difference_type n = 1 );
(since C++11)
(until C++17)
template< class BidirIt >

constexpr BidirIt prev(
  BidirIt it,

  typename std::iterator_traits<BidirIt>::difference_type n = 1 );
(since C++17)

Return the nth predecessor of iterator it.

Parameters

it - an iterator
n - number of elements it should be descended
Type requirements
-
BidirIt must meet the requirements of LegacyBidirectionalIterator.

Return value

The nth predecessor of iterator it.

Complexity

Linear.

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

Possible implementation

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

Notes

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, --c.end() does not compile, while std::prev(c.end()) does.

Example

#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';
}

Output:

1

See also

(C++11)
increment an iterator
(function template)
advances an iterator by given distance
(function template)