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std::prev

From cppreference.com
< cpp‎ | iterator
 
 
Iterator library
Iterator concepts
Iterator primitives
Iterator adaptors
Stream iterators
Iterator customization points
Iterator operations
prev
(C++11)
(C++11)
Range access
(C++11)(C++14)
(C++11)(C++14)
(C++17)(C++20)
(C++14)(C++14)
(C++14)(C++14)
(C++17)
(C++17)
 
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.

Contents

[edit] Parameters

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

[edit] Return value

The nth predecessor of iterator it.

[edit] Complexity

Linear.

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

[edit] Possible implementation

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

[edit] 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.

[edit] 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

[edit] See also

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