Как запрограммировать экспоненту

#include <iostream>
#include <iterator>
#include <stdexcept>
#include <numeric>
 
class ExpIterator;
 
ExpIterator begin( const double );
ExpIterator end( const double );
 
class ExpIterator : public std::iterator<std::input_iterator_tag, double> {
public:
  friend ExpIterator begin( const double );
  friend ExpIterator end( const double );
  ExpIterator( const ExpIterator& it ){
    x = it.x;
    dPrecision = it.dPrecision;
    dTemp = it.dTemp;
    bEnd = it.bEnd;
    uStep = it.uStep;
  }
  ExpIterator& operator++() {
    if( bEnd ){
      throw std::out_of_range( "bad idea ++ end iterator" );
    }else{
      dTemp *= x/uStep;
      ++uStep;
    }
    return *this;
  }
  ExpIterator operator++(int){ ExpIterator tmp(*this); operator++(); return tmp; }
  bool operator==(const ExpIterator& rhs) {
    if( !rhs.bEnd && !bEnd ){
      return( uStep == rhs.uStep );
    }else{
      return( bEnd ? rhs.dTemp < dPrecision : dTemp < rhs.dPrecision );
    }
  }
  bool operator!=(const ExpIterator& rhs) {return !operator==(rhs);}
  double operator*() {return dTemp;}
private:
  ExpIterator( const double _x, const double _dPrecision = 0, const bool _bEnd = false ) : 
    x(_x), dPrecision(_dPrecision), uStep(1), bEnd(_bEnd)  {
    if( !bEnd ){ dTemp = 1.0; }
  }
  double x, dPrecision, dTemp;
  unsigned uStep;
  bool bEnd;
};
 
ExpIterator begin( const double x ){ return ExpIterator( x ); }
ExpIterator end( const double dPrecision ){ return ExpIterator( 0, dPrecision, true ); }
 
int main(){
  ExpIterator it = begin( 5.0 ), end_it = end( 1e-5 );
  std::cout << std::accumulate( it, end_it, .0 );
  return 0;
}