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                       Standard C++ Library
             Copyright 1998, Rogue Wave Software, Inc.


NAME
     copy, copy_backward

      - Copies a range of elements.



SYNOPSIS
     #include <algorithm>
     template <class InputIterator, class OutputIterator>
     OutputIterator copy(InputIterator first,
                         InputIterator last,
                         OutputIterator result);
     template <class BidirectionalIterator1,
              class BidirectionalIterator2>
     BidirectionalIterator2
      copy_backward(BidirectionalIterator1 first,
                    BidirectionalIterator1 last,
                    BidirectionalIterator2 result);



DESCRIPTION
     The copy algorithm copies values from the range specified by
     [first,  last)  to  the range specified by [result, result +
     (last - first)). copy can be used to copy  values  from  one
     container to another, or to copy values from one location in
     a container to another location in the  same  container,  as
     long  as  result  is not within the range [first-last). copy
     returns result +  (last  -  first).  For  each  non-negative
     integer  n  <  (last  - first), copy assigns *(first + n) to
     *(result + n). The result of copy is undefined if result  is
     in the range [first, last).

     Unless result is an insert iterator, copy  assumes  that  at
     least  as  many  elements  follow result as are in the range
     [first, last).

     The copy_backward algorithm copies  elements  in  the  range
     specified  by  [first,  last)  into  the  range specified by
     [result - (last - first), result), starting from the end  of
     the  sequence (last-1) and progressing to the front (first).
     Note that copy_backward does not reverse the  order  of  the
     elements,   it   simply  reverses  the  order  of  transfer.
     copy_backward returns result - (last -  first).  You  should
     use  copy_backward instead of copy when last is in the range
     [result - (last - first), result). For each positive integer
     n  <=  (last  - first), copy_backward assigns *(last - n) to
     *(result - n). The result of copy_backward is  undefined  if
     result is in the range [first, last).

     Unless result is an insert iterator,  copy_backward  assumes
     that  there are at least as many elements ahead of result as
     are in the range [first, last).



COMPLEXITY
     Both copy_and copy_backward perform  exactly  last  -  first
     assignments.



EXAMPLE
        //
        // stdlib/examples/manual/copyex.cpp
        //
      #include <algorithm>
      #include <vector>
      #include <iostream>
     using namespace std;

     int main()
      {
       int d1[4] = {1,2,3,4};
       int d2[4] = {5,6,7,8};

        // Set up three vectors
        //
       vector<int> v1(d1,d1 + 4), v2(d2,d2 + 4), v3(d2,d2 + 4);
        //
        // Set up one empty vector
        //
       vector<int> v4;
        //
        // Copy v1 to v2
        //
       copy(v1.begin(),v1.end(),v2.begin());
        //
        // Copy backwards v1 to v3
        //
       copy_backward(v1.begin(),v1.end(),v3.end());
        //
        // Use insert iterator to copy into empty vector
        //
       copy(v1.begin(),v1.end(),back_inserter(v4));
        //
        // Copy all four to cout
        //
       ostream_iterator<int,char> out(cout," ");
       copy(v1.begin(),v1.end(),out);
       cout << endl;
       copy(v2.begin(),v2.end(),out);
       cout << endl;
       copy(v3.begin(),v3.end(),out);
       cout << endl;
       copy(v4.begin(),v4.end(),out);
       cout << endl;

       return 0;
      }

     Program Output



     1 2 3 4
     1 2 3 4
     1 2 3 4
     1 2 3 4



WARNINGS
     If your compiler does not support default  template  parame-
     ters,  then you always need to supply the Allocator template
     argument. For instance, you have to write:

     vector <int, allocator<int> >

     instead of:

     vector <int>

     If your compiler does not support namespaces,  then  you  do
     not need the using declaration for std.

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