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search

Category: algorithms Component type: function

Prototype

Search is an overloaded name; there are actually two search functions.
template <class ForwardIterator1, class ForwardIterator2>
ForwardIterator1 search(ForwardIterator1 first1, ForwardIterator1 last1,
                        ForwardIterator2 first2, ForwardIterator2 last2);

template <class ForwardIterator1, class ForwardIterator2, class BinaryPredicate>
ForwardIterator1 search(ForwardIterator1 first1, ForwardIterator1 last1,
                        ForwardIterator2 first2, ForwardIterator2 last2,
                        BinaryPredicate binary_pred);

Description

Search finds a subsequence within the range [first1, last1) that is identical to [first2, last2) when compared element-by-element. It returns an iterator pointing to the beginning of that subsequence, or else last1 if no such subsequence exists. The two versions of search differ in how they determine whether two elements are the same: the first uses operator==, and the second uses the user-supplied function object binary_pred.

The first version of search returns the first iterator i in the range [first1, last1 - (last2 - first2)) [1] such that, for every iterator j in the range [first2, last2), *(i + (j - first2)) == *j. The second version returns the first 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).

Definition

Defined in the standard header algorithm, and in the nonstandard backward-compatibility header algo.h.

Requirements on types

For the first version: For the second version:

Preconditions

Complexity

Worst case behavior is quadratic: at most (last1 - first1) * (last2 - first2) comparisons. This worst case, however, is rare. Average complexity is linear.

Example

  const char S1[] = "Hello, world!";
  const char S2[] = "world";
  const int N1 = sizeof(S1) - 1;
  const int N2 = sizeof(S2) - 1;

  const char* p = search(S1, S1 + N1, S2, S2 + N2);
  printf("Found subsequence \"%s\" at character %d of sequence \"%s\".\n",
         S2, p - S1, S1);

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 search with arguments such that last1 - first1 is less than last2 - first2, but such a search will always fail.

See also

find, find_if, find_end, search_n, mismatch, equal


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