std::search_n with example in C++

Prerequisite: std::search
std::search_n is an STL algorithm defined inside the header file , which is used to search whether a given element satisfies a predicate (equality if no such predicate is defined ) a given no. of times consecutively with the container elements.

It searches the range [first, last) for a sequence of count elements, each comparing equal to a given value (version 1) or satisfying a predicate (version 2).

The function returns an iterator to the first of such elements, or an iterator to the last element of the container, if no such sequence is found.

The two versions of std::search_n are defined as below –

1. For comparing elements using ==:

   Syntax:

    ForwardIterator search_n (ForwardIterator first, ForwardIterator last,
                              Size count, const T& val);
   
   first: 
   Forward iterator to beginning of the container to be searched into.
   last: 
   Forward iterator to end of the container to be searched into.
   count: 
   Minimum number of successive elements to match.
   Size shall be (convertible to) an integral type.
   val: Individual value to be compared.
   Returns: 
   It returns an iterator to the first element of the sequence.
   If no such sequence is found, the function returns last.
   

   // C++ program to demonstrate the use of std::search_n
   
   #include <iostream>
   #include <vector>
   #include <algorithm>
   using namespace std;
   int main()
   {
       int i, j;
   
       // Declaring the sequence to be searched into
       vector<int> v1 = { 1, 2, 3, 4, 5, 3, 3, 6, 7 };
   
       // Declaring the value to be searched for
       int v2 = 3;
   
       // Declaring an iterator for storing the returning pointer
       vector<int>::iterator i1;
   
       // Using std::search_n and storing the result in
       // iterator i1
       i1 = std::search_n(v1.begin(), v1.end(), 2, v2);
   
       // checking if iterator i1 contains end pointer of v1 or not
       if (i1 != v1.end()) {
           cout << "v2 is present consecutively 2 times at index "
                << (i1 - v1.begin());
       } else {
           cout << "v2 is not present consecutively 2 times in "
                << "vector v1";
       }
   
       return 0;
   }

   Output:

   v2 is present consecutively 2 times at index 5
   

2. For comparing element using a predicate:
   Syntax:

   ForwardIterator search_n ( ForwardIterator first, ForwardIterator last,
                              Size count, const T& val, BinaryPredicate pred );
   
   All the arguments are same as previous template, just one more argument is added
   
   pred: Binary function that accepts two arguments 
   (one element from the sequence as first, and val as second),
    and returns a value convertible to bool. The value returned indicates whether 
   the element is considered a 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.
   
   Returns:
   It also returns value as per the previous version, i.e., 
   an iterator to the first element of the sequence, 
   satisfying a condition with respect to a given value.
   If no such sequence is found, the function returns last.

   // C++ program to demonstrate the use of std::search_n
   // with binary predicate
   #include <iostream>
   #include <vector>
   #include <algorithm>
   using namespace std;
   
   // Defining the BinaryPredicate function
   bool pred(int i, int j)
   {
       if (i == j) {
           return 1;
       } else {
           return 0;
       }
   }
   
   int main()
   {
       int i, j;
   
       // Declaring the sequence to be searched into
       vector<int> v1 = { 1, 2, 3, 4, 5, 3, 3, 6, 7 };
   
       // Declaring the value to be compared to v1 based
       // on a given predicate
       int v2 = 3;
   
       // Declaring an iterator for storing the returning pointer
       vector<int>::iterator i1;
   
       // Using std::search_n and storing the result in
       // iterator i1
       i1 = std::search_n(v1.begin(), v1.end(), 2, v2, pred);
   
       // checking if iterator i1 contains end pointer of v1 or not
       if (i1 != v1.end()) {
           cout << "v2 is present consecutively 2 times at index "
                << (i1 - v1.begin());
       } else {
           cout << "v2 is not present consecutively 2 times "
                << "in vector v1";
       }
   
       return 0;
   }

   Output:

   v2 is present consecutively 2 times at index 5
   

If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.