C++ Algorithm Library - find_if_not() Function

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Description

The C++ function std::algorithm::find_if_not() finds the last occurrence of the element that satisfies the condition. It uses unary predicate to specify condition.

Declaration

Following is the declaration for std::algorithm::find_if_not() function form std::algorithm header.

C++11

template <class InputIterator, class UnaryPredicate>
InputIterator find_if_not(InputIterator first, InputIterator last, UnaryPredicate pred);

Parameters

• first − Input iterator to the initial position.

• last − Final iterator to the final position.

• pred − A unary predicate which accepts one argument and returns bool.

Return value

returns an iterator to the first element in the range (first,last) for which unary predicate returns false. If no such element is found, the function returns last.

Exceptions

Throws exception if either predicate or an operation on an iterator throws exception.

Please note that invalid parameters cause undefined behavior.

Time complexity

Linear.

Example

The following example shows the usage of std::algorithm::find_if_not() function.

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

bool unary_pred(int n) {
   return ((n % 2) == 0);
}

int main(void) {
   vector<int> v = {2, 4, 61, 8, 10};
   auto it = find_if_not(v.begin(), v.end(), unary_pred);

   if (it != end(v))
      cout << "First odd number is " << *it << endl;

   v = {2, 4, 6, 8, 10};

   it = find_if_not(v.begin(), v.end(), unary_pred);

   if (it == end(v))
      cout << "Only enven elements present in the sequence." << endl;

   return 0;
}

Let us compile and run the above program, this will produce the following result −

First odd number is 61
Only enven elements present in the sequence.