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unique_copy

Copies elements from a source range into a destination range except for the duplicate elements that are adjacent to each other.

template<class InputIterator, class OutputIterator>
   OutputIterator unique_copy(
      InputIterator _First, 
      InputIterator _Last, 
      OutputIterator _Result
   );
template<class InputIterator, class OutputIterator, class BinaryPredicate>
   OutputIterator unique_copy(
      InputIterator _First, 
      InputIterator _Last, 
      OutputIterator _Result,
      BinaryPredicate _Comp,
   );

Parameters

_First
A forward iterator addressing the position of the first element in the source range to be copied.
_Last
A forward iterator addressing the position one past the final element in the source range to be copied.
_Result
An output iterator addressing the position of the first element in the destination range that is receiving the copy with consecutive duplicates removed.
_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.

Return Value

An output iterator addressing the position one past the final element in the destination range that is receiving the copy with consecutive duplicates removed.

Remarks

Both forms of the algorithm remove the second duplicate of a consecutive pair of equal elements.

The operation of the algorithm is stable so that the relative order of the undeleted elements is not changed.

The ranges referenced must be valid; all pointers must be dereferenceable and within a sequence the last position is reachable from the first by incrementation.

The complexity is linear, requiring (_Last – _First) comparisons.

Example

// alg_unique_copy.cpp
// compile with: /EHsc
#include <vector>
#include <algorithm>
#include <functional>
#include <iostream>
#include <ostream>

using namespace std;

// Return whether modulus of elem1 is equal to modulus of elem2
bool mod_equal ( int elem1, int elem2 )
{
   if ( elem1 < 0 ) 
      elem1 = - elem1;
   if ( elem2 < 0 ) 
      elem2 = - elem2;
   return elem1 == elem2;
};

int main( )
{
   vector <int> v1;
   vector <int>::iterator v1_Iter1, v1_Iter2,
         v1_NewEnd1, v1_NewEnd2;

   int i;
   for ( i = 0 ; i <= 1 ; i++ )
   {
      v1.push_back( 5 );
      v1.push_back( -5 );
   }

   int ii;
   for ( ii = 0 ; ii <= 2 ; ii++ )
   {
      v1.push_back( 4 );
   }
   v1.push_back( 7 );

   int iii;
   for ( iii = 0 ; iii <= 5 ; iii++ )
   {
      v1.push_back( 10 );
   }
   
   cout << "Vector v1 is\n ( " ;
   for ( v1_Iter1 = v1.begin( ) ; v1_Iter1 != v1.end( ) ; v1_Iter1++ )
      cout << *v1_Iter1 << " ";
   cout << ")." << endl;

   // Copy first half to second, removing consecutive duplicates
   v1_NewEnd1 = unique_copy ( v1.begin ( ) , v1.begin ( ) + 8, v1.begin ( ) + 8 );

   cout << "Copying the first half of the vector to the second half\n "
        << "while removing adjacent duplicates gives\n ( " ;
   for ( v1_Iter1 = v1.begin( ) ; v1_Iter1 != v1_NewEnd1 ; v1_Iter1++ )
      cout << *v1_Iter1 << " ";
   cout << ")." << endl;

   int iv;
   for ( iv = 0 ; iv <= 7 ; iv++ )
   {
      v1.push_back( 10 );
   }

   // Remove consecutive duplicates under the binary prediate mod_equals
   v1_NewEnd2 = unique_copy ( v1.begin ( ) , v1.begin ( ) + 14, 
      v1.begin ( ) + 14 , mod_equal );

   cout << "Copying the first half of the vector to the second half\n "
        << " removing adjacent duplicates under mod_equals gives\n ( " ;
   for ( v1_Iter2 = v1.begin( ) ; v1_Iter2 != v1_NewEnd2 ; v1_Iter2++ )
      cout << *v1_Iter2 << " ";
   cout << ")." << endl;
}

Output

Vector v1 is
 ( 5 -5 5 -5 4 4 4 7 10 10 10 10 10 10 ).
Copying the first half of the vector to the second half
 while removing adjacent duplicates gives
 ( 5 -5 5 -5 4 4 4 7 5 -5 5 -5 4 7 ).
Copying the first half of the vector to the second half
  removing adjacent duplicates under mod_equals gives
 ( 5 -5 5 -5 4 4 4 7 5 -5 5 -5 4 7 5 4 7 5 4 7 ).

See Also

<algorithm> Members

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