C++/Deque/deque iterator

Define iterator for deque

<source lang="cpp">

// Demonstrating the generic find algorithm with a deque

1. include <iostream>
2. include <cassert>
3. include <deque>
4. include <algorithm> // For find

using namespace std;

int main() {

 int x[5] = {1, 2, 3, 4, 5};
 deque<int> deque1(&x[0], &x[5]);
 // Search for the first occurrence of the letter e:
 deque<int>::iterator i;
 cout.precision(10);
 for (i = deque1.begin(); i != deque1.end(); ++i)
   cout << *i << endl;
 return 0;

} /* 1 2 3 4 5

*/        
 </source>

Use iterator and reverse_iterator with deque

<source lang="cpp">

/* The following code example is taken from the book

* "The C++ Standard Library - A Tutorial and Reference"
* by Nicolai M. Josuttis, Addison-Wesley, 1999
*
* (C) Copyright Nicolai M. Josuttis 1999.
* Permission to copy, use, modify, sell and distribute this software
* is granted provided this copyright notice appears in all copies.
* This software is provided "as is" without express or implied
* warranty, and with no claim as to its suitability for any purpose.
*/

1. include <iostream>
2. include <deque>
3. include <algorithm>

using namespace std; void print (int elem) {

   cout << elem << " ";

} int main() {

   deque<int> coll;
   // insert elements from 1 to 9
   for (int i=1; i<=9; ++i) {
       coll.push_back(i);
   }
   // find position of element with value 2
   deque<int>::iterator pos1;
   pos1 = find (coll.begin(), coll.end(),    // range
                2);                          // value
   // find position of element with value 7
   deque<int>::iterator pos2;
   pos2 = find (coll.begin(), coll.end(),    // range
                7);                          // value
   // print all elements in range [pos1,pos2)
   for_each (pos1, pos2,     // range
             print);         // operation
   cout << endl;
   // convert iterators to reverse iterators
   deque<int>::reverse_iterator rpos1(pos1);
   deque<int>::reverse_iterator rpos2(pos2);
   // print all elements in range [pos1,pos2) in reverse order
   for_each (rpos2, rpos1,   // range
             print);         // operation
   cout << endl;

} /* 2 3 4 5 6 6 5 4 3 2

*/
       
 </source>