Re: MSDN for_each sample

From:
David Wilkinson <no-reply@effisols.com>
Newsgroups:
microsoft.public.vc.language
Date:
Mon, 03 Dec 2007 06:35:35 -0500
Message-ID:
<e1aodCaNIHA.3400@TK2MSFTNGP03.phx.gbl>
George wrote:

Hello everyone,

I do not know why in for_each sample of MSDN, operator double() is invoked.
Which statement triggers operator double()?

http://msdn2.microsoft.com/en-us/library/e5sk9w9k(VS.71).aspx

[Code]
#include <vector>
#include <algorithm>
#include <iostream>

// The function object to determine the average
class Average
{
private:
   long num; // The number of elements
   long sum; // The sum of the elements
public:
   // Constructor initializes the value to multiply by
   Average ( ) : num ( 0 ) , sum ( 0 )
   {
   }

   // The function call to process the next elment
   void operator ( ) ( int elem ) \
   {
      num++; // Increment the element count
      sum += elem; // Add the value to the partial sum
   }

   // return Average
   operator double ( )
   {
      return static_cast <double> (sum) /
      static_cast <double> (num);
   }
};

int main( )
{
   using namespace std;
   vector <int> v1;
   vector <int>::iterator Iter1;

   // Constructing vector v1
   int i;
   for ( i = -4 ; i <= 2 ; i++ )
   {
      v1.push_back( i );
   }

   // The local state of a function object can accumulate
   // information about a sequence of actions that the
   // return value can make available, here the Average
   double avemod2 = for_each ( v1.begin ( ) , v1.end ( ) ,
      Average ( ) );
   cout << "The average of the elements of v1 is:\n Average ( v1mod2 ) = "
        << avemod2 << "." << endl;
}
[/Code]


George:

std::for_each returns a copy of the (possibly modified) unary function
(third argument).

So the statement

double avemod2 = for_each ( v1.begin( ), v1.end( ), Average( ) );

requires conversion of Average to double.

--
David Wilkinson
Visual C++ MVP

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