"clicwar" <clic...@gmail.com> wrote in message
news:1185144847.041900.64380@q75g2000hsh.googlegroups.com...
Jim, did you mean something like this?
#include <iostream>
#include <string>
using namespace std;
class Vector {
private:
float x,y;
public:
Vector(float u, float v);
Vector();
Vector operator+ (const Vector &a) const;
Vector(const Vector &source);
void Show();
};
void Vector::Show() {
cout <<"(" << x <<"," <<y <<")";
}
Vector::Vector(float u, float v) {
x=u; y=v;
}
Vector::Vector() {
x=0; y=0;
}
Vector::Vector(const Vector &source) {
x = (source.x)*2 ; y = (source.y)*2 ;
}
Vector Vector::operator+ (const Vector &a) const {
Vector temp;
temp.x = x + a.x;
temp.y = y + a.y;
return (temp);
}
int main() {
Vector a(3,1), b(5,2), c, d;
c = (a+b);
d = (a.operator+ (b));
cout << "Data members of the vector c: ";
c.Show();
Vector e((a+b));
cout <<endl << "Data members of the vector e: ";
e.Show();
return 0;
}
Did you use g++ ? Because now i'm getting (8,3) for both c and e.
Thanks for the patience and for the (void) tip.
Here, your "copy constructor" doesn't make an actual copy. Are you
sure
you
want to do that?
IV, how can i improve this? I came from a little C background and i
just learned how to do operator overloading.
On 22 jul, 18:59, I V <ivle...@gmail.com> wrote:
On Sun, 22 Jul 2007 19:11:46 +0000, clicwar wrote:
A simple program with operator overloading and copy constructor:
[...]
Vector::Vector(Vector &source) {
x = (source.x)*2 ; y = (source.y)*2 ;
}
Here, your "copy constructor" doesn't make an actual copy. Are you
sure
you
want to do that? It seems to me to be likely to cause confusion.
Indeed,
I
think this is why you're not getting the result you expect:
Vector a(3,1), b(5,2), c, d;
c = a+b;
Now, this calls a.operator+(b), which invokes the copy constructor on
b;
so a.operator+ gets passed (10, 4); it then adds this to (3, 1),
giving
(13, 5), and returns this, which itself invokes the copy constructor,
giving (26, 10). However, I think this last use of the copy
constructor,
in the return, is optional; so the results of calling operator+ depend
on
whether or not the compiler decides to optimize out the copy
constructor.
This doesn't seem like a very good state of affairs.
Here's your program cleaned up a bit and extened a little.
#include <iostream>
#include <string>
class Vector
{
private:
float x,y;
public:
Vector(float u, float v);
Vector operator+ (const Vector &a) const;
Vector operator* (const float a ) const;
Vector operator* (const Vector &b) const;
Vector(const Vector &source);
void Show();
};
void Vector::Show()
{
std::cout <<"(" << x <<"," <<y <<")";
}
Vector::Vector(float u = 0, float v = 0): x(u), y(v)
{
}
Vector::Vector(const Vector &source)
{
x = source.x;
y = source.y;
}
Vector Vector::operator+ (const Vector &a) const
{
Vector temp;
temp.x = x + a.x;
temp.y = y + a.y;
return temp;
}
Vector Vector::operator* (float a) const
{
Vector temp;
temp.x = x * a;
temp.y = y * a;
return temp;
}
Vector Vector::operator* (const Vector& a ) const
{
Vector temp;
temp.x = x * a.x;
temp.y = y * a.y;
return temp;
}
int main()
{
Vector a(3,1), b(5,2), c, d;
c = a+b;
d = (a.operator+ (b));
std::cout << "Data members of the vector c: ";
c.Show();
Vector e((a+b));
std::cout << std::endl << "Data members of the vector e: ";
e.Show();
std::cout << "\na * 2 = ";
(a * 2).Show();
std::cout << "\na * b = ";
(a * b).Show();
return 0;
}
All a copy constructor should do is copy. x = x, y = y. Nothing more.
If
you want to mulitply, have an operator*
The multiplication of the two Vectors (Vector * Vector) is arbitary. A
lot
of times multiplying two vectors would return the dot product. It dpends
on
what you want it to do.