Expert: vijayan - 3/15/2010

Question
QUESTION: Hello vijayan,

here is a question i tried to solve, could you check my solution and let me know if i am close and if any error, fix it abit for me?

question:

To define a point in a two-dimensional coordinate system, it is most common to use the form (x,y), called rectangular coordinates. Construct a class Point which describes points in this way. Write appropriate constructors and a member function which calculates the distance between a given point and another. The distance between two points (x1,y1) and (x2,y2) is given by the formula: d = √((x1-y1)2+(x2-y2)2)

my solution
===============================================================
#include<iostream>
#include<math.h>

using namespace std;
class point{
     int x, y;
     point(int a = 0, int b=0);
     int cal_Distance(point &b);
     };

point ::point(): x(a),y(b){}
int  point::cal_Distance(point &b)
{
  int d;
   d = sqrt((x-y)^2)+(b.x-b.y)^2);
   return d;
}

int main()
{
   point a(2,1);
   point b(3,1);
   float d;
  d =  (float)a.call_Distance(point &b);
   cout<<d<<endl;
   getche();
}
===============================================================

let me know of my errors by commenting them when correcting this code..

thanks..

ANSWER: class point
{
    int x, y;

    //add access specifier
    public:

    point(int a = 0, int b=0);

    // not const-correct
    // int cal_Distance(point &b);
    // modify as (return a double, add const)
    double cal_Distance( const point &b ) const ;
    };

point::point( int a, int b ): x(a), y(b){} // *** parameters added

double point::cal_Distance( const point &b ) const
{
 int x2 = b.x ;
 int y2 = b.y ;

 return sqrt( ( (x-y) * (x-y) ) + ( (x2-y2) * (x2-y2) ) ) ; // *** corrected
}

int main()
{
  point a(2,1);
  point b(3,1);
  double d;
  d =  (float)a.cal_Distance(b); // *** corrected
  cout<<d<<endl;
  //getche(); // *** deleted
}


---------- FOLLOW-UP ----------

QUESTION: Hello vijayan,

thnks, i saw my errors and i corrected them. here is a full version of the different codes:

==========================================================================
using namespace std;
class point{
     int x, y;
     public:
     point(int a = 0, int b=0);
     double cal_Distance(point &b) const;
     };

point ::point(int a, int b): x(a),y(b){}
double  point::cal_Distance(point &b) const
{
  int x2 = b.x;
  int y2 = b.y;
 
return sqrt( ( (x-y) * (x-y) ) + ( (x2-y2) * (x2-y2) ) ) ;
 
}

int main()
{
   point a(2,1);
   point b(3,1);
double  d;
  d =  a.cal_Distance(b);
   cout<<d<<endl;
   getche();
}
=======================================================================
my next task now is a conversion from polar coordinates to ordinary rectangular coordinates and this can be effected with the formula
X = r.cosϴ
Y = r.sinϴ
First define a class RPoint, where the points are defined using polar coordinates(that i have done using the class Pointer. just rename the Pointer class as RPoint), then write a type conversion constructor for class Point which converts RPoint to a Point.

i got no clue how to write a type conversion constructor class. so, i am counting on your expertise to define this and i can learn how to do that.. you could give a small work frame around what i have done so that i am inline with your way.

Answer
> i got no clue how to write a type conversion constructor class

A type conversion constructor for a class is a non-explicit constructor in the destination type that can be invoked with a single argument of the source type. In your case, the constructor would be:

Point::Point( const RPoint& rpt ) ;