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count

Count appearances of value in range

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Contents

Definition
Interface
Description
Return Value
Complexity
Page Comments

Key Facts

Gyroscopic Couple: The rate of change of angular momentum ( $\inline \tau$ ) = $\inline I\omega\Omega$ (In the limit).

$\inline I$ = Moment of Inertia.
$\inline \omega$ = Angular velocity
$\inline \Omega$ = 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 count() algorithm is defined in the standard header <algorithm> and in the nonstandard backward-compatibility header <algo.h>.

Interface

#include <algorithm>
template < class InputIterator, class Type>
   typename iterator_traits<InputIterator>::difference_type count(
      InputIterator first, 
      InputIterator last, 
      const Type& val
   );

Parameters:

Parameter	Description
first	An input iterator addressing the position of the first element in the range to be traversed
last	An input iterator addressing the position one past the final element in the range to be traversed
val	The value of the elements to be counted

Description

Count the number of values in an input range [first,last) that equal val.

Return Value

Count algorithm returns the number of iterators i in [first, last) such that *i == value.

Complexity

The complexity is linear. Exactly last - first comparisons.

Example:

Example - count algorithm

Problem

This short program illustrates the functionality of count() algorithm.

Workings

#include <vector>
#include <algorithm>
#include <iostream>
 
using namespace std;
 
int main()
{
  vector<int> v1;
  vector<int>::iterator Iter;
 
  v1.push_back(10);
  v1.push_back(20);
  v1.push_back(10);
  v1.push_back(40);
  v1.push_back(10);
 
  cout <<"v1 = ( " ;
  for (Iter = v1.begin(); Iter != v1.end(); Iter++)
      cout <<*Iter<<" ";
  cout <<")"<<endl;
 
  vector<int>::iterator::difference_type result;
  result = count(v1.begin(), v1.end(), 10);
  cout <<"The number of 10s in v2 is: "<<result<<"."<<endl;
 
  return 0;
}

Solution

Output:

v1 = ( 10 20 10 40 10 )
The number of 10s in v2 is: 3.