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std::prev

Defined in header <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)

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 BidirectionalIterator.

Return value

The nth predecessor of iterator it.

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: