std::find_end

equality (1)	template <class ForwardIterator1, class ForwardIterator2> ForwardIterator1 find_end (ForwardIterator1 first1, ForwardIterator1 last1, ForwardIterator2 first2, ForwardIterator2 last2);
predicate (2)	template <class ForwardIterator1, class ForwardIterator2, class BinaryPredicate> ForwardIterator1 find_end (ForwardIterator1 first1, ForwardIterator1 last1, ForwardIterator2 first2, ForwardIterator2 last2, BinaryPredicate pred);

equality (1)

template <class ForwardIterator1, class ForwardIterator2>   ForwardIterator1 find_end (ForwardIterator1 first1, ForwardIterator1 last1,                              ForwardIterator2 first2, ForwardIterator2 last2);

predicate (2)

template <class ForwardIterator1, class ForwardIterator2, class BinaryPredicate>   ForwardIterator1 find_end (ForwardIterator1 first1, ForwardIterator1 last1,                              ForwardIterator2 first2, ForwardIterator2 last2,                              BinaryPredicate pred);

Find last subsequence in range

Searches the range [first1,last1) for the last occurrence of the sequence defined by [first2,last2), and returns an iterator to its first element, or last1 if no occurrences are found.

The elements in both ranges are compared sequentially using operator== (or pred, in version (2)): A subsequence of [first1,last1) is considered a match only when this is true for all the elements of [first2,last2).

This function returns the last of such occurrences. For an algorithm that returns the first instead, see search.

The behavior of this function template is equivalent to:

template<class ForwardIterator1, class ForwardIterator2>
  ForwardIterator1 find_end (ForwardIterator1 first1, ForwardIterator1 last1,
                             ForwardIterator2 first2, ForwardIterator2 last2)
{
  if (first2==last2) return last1;  // specified in C++11

  ForwardIterator1 ret = last1;

  while (first1!=last1)
  {
    ForwardIterator1 it1 = first1;
    ForwardIterator2 it2 = first2;
    while (*it1==*it2) {    // or: while (pred(*it1,*it2)) for version (2)
        ++it1; ++it2;
        if (it2==last2) { ret=first1; break; }
        if (it1==last1) return ret;
    }
    ++first1;
  }
  return ret;
}

Parameters

first1, last1: Forward iterators to the initial and final positions of the searched sequence. The range used is [first1,last1), which contains all the elements between first1 and last1, including the element pointed by first1 but not the element pointed by last1.
first2, last2: Forward iterators to the initial and final positions of the sequence to be searched for. The range used is [first2,last2).
For (1), the elements in both ranges shall be of types comparable using operator== (with the elements of the first range as left-hand side operands, and those of the second as right-hand side operands).
pred: Binary function that accepts two elements as arguments (one of each of the two sequences, in the same order), and returns a value convertible to bool. The returned value indicates whether the elements are considered to match in the context of this function.
The function shall not modify any of its arguments.
This can either be a function pointer or a function object.

Return value

An iterator to the first element of the last occurrence of [first2,last2) in [first1,last1).
If the sequence is not found, the function returns last1.

If [first2,last2) is an empty range, the result is unspecified.

If [first2,last2) is an empty range, the function returns last1.

Example

// find_end example
#include <iostream>     // std::cout
#include <algorithm>    // std::find_end
#include <vector>       // std::vector

bool myfunction (int i, int j) {
  return (i==j);
}

int main () {
  int myints[] = {1,2,3,4,5,1,2,3,4,5};
  std::vector<int> haystack (myints,myints+10);

  int needle1[] = {1,2,3};

  // using default comparison:
  std::vector<int>::iterator it;
  it = std::find_end (haystack.begin(), haystack.end(), needle1, needle1+3);

  if (it!=haystack.end())
    std::cout << "needle1 last found at position " << (it-haystack.begin()) << '\n';

  int needle2[] = {4,5,1};

  // using predicate comparison:
  it = std::find_end (haystack.begin(), haystack.end(), needle2, needle2+3, myfunction);

  if (it!=haystack.end())
    std::cout << "needle2 last found at position " << (it-haystack.begin()) << '\n';

  return 0;
}

Output:

needle1 last found at position 5
needle2 last found at position 3

Complexity

Up to linear in count2*(1+count1-count2), where countX is the distance between firstX and lastX: Compares elements until the last matching subsequence is found.

Data races

Some (or all) of the objects in both ranges are accessed (possibly more than once).

Exceptions

Throws if any element comparison (or call to pred) throws or if any of the operations on iterators throws.
Note that invalid arguments cause undefined behavior.