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find_end
Prototypefind_end is an overloaded name; there are actually two find_end functions.template <class ForwardIterator1, class ForwardIterator2> ForwardIterator1 find_end(ForwardIterator1 first1, ForwardIterator1 last1, ForwardIterator2 first2, ForwardIterator2 last2); template <class ForwardIterator1, class ForwardIterator2, class BinaryPredicate> ForwardIterator1 find_end(ForwardIterator1 first1, ForwardIterator1 last1, ForwardIterator2 first2, ForwardIterator2 last2, BinaryPredicate comp); DescriptionFind_end is misnamed: it is much more similar to search than to find, and a more accurate name would have been search_end.Like search, find_end attempts to find a subsequence within the range [first1, last1) that is identical to [first2, last2). The difference is that while search finds the first such subsequence, find_end finds the last such subsequence. Find_end returns an iterator pointing to the beginning of that subsequence; if no such subsequence exists, it returns last1. The two versions of find_end differ in how they determine whether two elements are the same: the first uses operator==, and the second uses the usersupplied function object comp. The first version of find_end returns the last iterator i in the range [first1, last1  (last2  first2)) such that, for every iterator j in the range [first2, last2), *(i + (j  first2)) == *j. The second version of find_end returns the last iterator i in [first1, last1  (last2  first2)) such that, for every iterator j in [first2, last2), binary_pred(*(i + (j  first2)), *j) is true. These conditions simply mean that every element in the subrange beginning with i must be the same as the corresponding element in [first2, last2). DefinitionDefined in the standard header algorithm, and in the nonstandard backwardcompatibility header algo.h.Requirements on typesFor the first version:
Preconditions
ComplexityThe number of comparisons is proportional to (last1  first1) * (last2  first2). If both ForwardIterator1 and ForwardIterator2 are models of Bidirectional Iterator, then the average complexity is linear and the worst case is at most (last1  first1) * (last2  first2) comparisons.Exampleint main() { char* s = "executable.exe"; char* suffix = "exe"; const int N = strlen(s); const int N_suf = strlen(suffix); char* location = find_end(s, s + N, suffix, suffix + N_suf); if (location != s + N) { cout << "Found a match for " << suffix << " within " << s << endl; cout << s << endl; int i; for (i = 0; i < (location  s); ++i) cout << ' '; for (i = 0; i < N_suf; ++i) cout << '^'; cout << endl; } else cout << "No match for " << suffix << " within " << s << endl; } Notes[1] The reason that this range is [first1, last1  (last2  first2)), instead of simply [first1, last1), is that we are looking for a subsequence that is equal to the complete sequence [first2, last2). An iterator i can't be the beginning of such a subsequence unless last1  i is greater than or equal to last2  first2. Note the implication of this: you may call find_end with arguments such that last1  first1 is less than last2  first2, but such a search will always fail. See alsosearchCopyright © 1999 Silicon Graphics, Inc. All Rights Reserved. TrademarkInformation
