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# find_first_of

Searches the first of several possible elements View version details

### Key Facts

Gyroscopic Couple: The rate of change of angular momentum ( ) = (In the limit).
• = Moment of Inertia.
• = Angular velocity
• = Angular velocity of precession.

Blaise Pascal (1623-1662) was a French mathematician, physicist, inventor, writer and Catholic philosopher.

Leonhard Euler (1707-1783) was a pioneering Swiss mathematician and physicist.

## Definition

The find_first_of() algorithm is defined in the standard header <algorithm> and in the nonstandard backward-compatibility header <algo.h>.

## Interface

#include <algorithm>
template < class ForwardIterator1, class ForwardIterator2 >
ForwardIterator1 find_first_of(
ForwardIterator1 first1,
ForwardIterator1 last1,
ForwardIterator2 first2,
ForwardIterator2 last2
);

template < class ForwardIterator1,
class ForwardIterator2,
class BinaryPredicate >
ForwardIterator1 find_first_of(
ForwardIterator1 first1,
ForwardIterator1 last1,
ForwardIterator2 first2,
ForwardIterator2 last2,
BinaryPredicate comp
);

Parameters:

Parameter Description
first1 A forward iterator addressing the position of the first element in the range to be searched
last1 A forward iterator addressing the position one past the final element in the range to be searched
first2 A forward iterator addressing the position of the first element in the range to be matched
last2 A forward iterator addressing the position one past the final element in the range to be matched
comp User-defined predicate function object that defines the condition to be satisfied if two elements are to be taken as equivalent. A binary predicate takes two arguments and returns true when satisfied and false when not satisfied

## Description

Find_first_of is similar to find, in that it performs linear search through a range of Input Iterators. The difference is that while find searches for one particular value, find_first_of searches for any of several values.

The two versions of find_first_of differ in how they compare elements for equality.

The first uses operator== and the second uses and arbitrary user-supplied function object comp.

## Return Value

The first version returns the first iterator i in [first1, last1) such that, for some iterator j in [first2, last2), *i == *j.

The second returns the first iterator i in [first1, last1) such that, for some iterator j in [first2, last2), comp(*i, *j) is true. As usual, both versions return last1 if no such iterator i exists.

## Complexity

At most, performs distance1*distance2 comparisons or applications of comp (where distanceX is the distance between firstX and lastX).

### References

Example:
##### Example - find_first_of algorithm
Problem
This program illustrates the use of the STL find_first_of() algorithm (default version) to find the first occurrence of any one of a range of integer values in a vector within another range of integer values, also in a vector.
Workings
#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int main()
{
int a1[] = {1, 2, 333, 3, 4, 5, 6, 444, 7, 8, 9, 10};
vector<int> v1(a1, a1+12);
cout <<"\nHere are the contents of v1:\n";
for (vector<int>::size_type i=0; i<v1.size(); i++)
cout <<v1.at(i)<<" ";

int a2[] = {333, 444, 555};
vector<int> v2(a2, a2+3);
cout <<"\nHere are the contents of v2:\n";
for (vector<int>::size_type i=0; i<v2.size(); i++)
cout <<v2.at(i)<<" ";

vector<int>::iterator p;

p = find_first_of(v1.begin(), v1.end(), v2.begin(), v2.end());
if (p != v1.end())
cout <<"\nThe first instance of a value from v2 occurring in v1 happens at location "
<<(int)(p-v1.begin()+1)<<".";
else
cout <<"\nNo instance of v2 was found in v1.";

p = find_first_of(p+1, v1.end(), v2.begin(), v2.end());
if (p != v1.end())
cout <<"\nThe next instance of a value from v2 occurring in v1 happens   at location "
<<(int)(p-v1.begin()+1) << ".";
else
cout <<"\nNo further instance of a value from v2 occurring in v1 was    found.";

p = find_first_of(p+1, v1.end(), v2.begin(), v2.end());
if (p != v1.end())
cout <<"\nThe next instance of a value from v2 occurring in v1 happens   at location "
<<(int)(p-v1.begin()+1)<<".";
else
cout <<"\nNo further instance of a value from v2 occurring in v1 was    found.";

return 0;
}
Solution
Output:

Here are the contents of v1:
1 2 333 3 4 5 6 444 7 8 9 10

Here are the contents of v2:
333 444 555

The first instance of a value from v2 occurring in v1 happens at location 3.

The next instance of a value from v2 occurring in v1 happens at location 8.

No further instance of a value from v2 occurring in v1 was found.
References