Use set_difference to get values from only one list : difference « Set Multiset « C++






  1. C++
  2. Set Multiset
  3. difference

Use set_difference to get values from only one list

  
#include <algorithm>
#include <functional>
#include <iomanip>
#include <iostream>
#include <list>
#include <string>
#include <vector>

using namespace std;

class PC
{
   public:
   enum part { keyboard, mouse, monitor };

   PC( part a_part = PC::keyboard, int id = 0 );
   bool operator<( const PC& rhs ) const;

   void print() const;

   private:
   part part_;
   int id_;
};

inline
PC::PC( part a_part, int id ) : part_( a_part ), id_( id ){}

inline bool PC::operator<( const PC& rhs ) const{  
   return id_ < rhs.id_; 
}

void PC::print() const {
   string component;
   if( part_ == keyboard )
      component = "keyboard";
   else if( part_ == mouse )
      component = "mouse";
   else
      component = "monitor";

   cout << "ID: " << setw( 8 ) << left << id_ << " PC: " << component << endl;
}

int main( )
{
   list<PC> listA;
   listA.push_back( PC( PC::keyboard, 3 ) );
   listA.push_back( PC( PC::mouse, 1 ) );
   listA.push_back( PC( PC::monitor, 9 ) );
   listA.push_back( PC( PC::keyboard, 2 ) );
   listA.push_back( PC( PC::monitor, 8 ) );

   list<PC> inspector_B( listA );
   inspector_B.front() = PC( PC::mouse, 6 );
   inspector_B.back() = PC( PC::monitor, 1 );

   // must sort before using set algorithms
   listA.sort();
   inspector_B.sort();

   vector<PC> result;
   // make vector large enough to hold all inspected parts
   result.resize( listA.size() + inspector_B.size() );

   cout << "\n\nlist B ONLY\n";
   vector<PC>::iterator the_end =set_difference( inspector_B.begin(), inspector_B.end(),listA.begin(), listA.end(), result.begin() );
   for_each( result.begin(), the_end, mem_fun_ref( &PC::print ) );

}








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1.Get the difference between two sets
2.Get the symmetric difference between two sets
3.set_difference and back_inserter
4.Use set_symmetric_difference() to get the difference between two lists
5.Create the symmetric difference between list1 and list2