Re: Total Newbie Backward Compat Question

Crazy c said:

<snip>

> Are you saying that if I use std::cout, I do not have to overload
> operator <<?

If you want std::cout to be able to display objects of your own definition

(as opposed to vanilla types such as int, double, etc), then you need to 
overload << within your class.

For example... ooh boy, how did I get into this? ... let's say you have a 
point class, which stores x and y co-ordinates, and you want std::cout to 
be able to display (x, y). You'd do something like this (once you've 
checked for replies to this article correcting my blunders!):

#include <iostream>

class point
{
  int x;
  int y;
public:
  point(int x_, int y_) : x(x_), y(y_) {}
  int getx() const { return x; }
  int gety() const { return y; }
  int setx(int x_) { int r = x; x = x_; return r; }
  int sety(int y_) { int r = y; y = y_; return r; }
};

/* Stolen and adapted from Stroustrup Special Edition p612 */
std::ostream &operator<<(std::ostream &s, const point &p)
{
  return s << '(' << p.getx() << ", " << p.gety() << ')';
}

int main()
{
  point p(3, 4);
  std::cout << "My point is " << p << ", okay?" << std::endl;
  p.setx(6);
  std::cout << "My point is now " << p << " instead." << std::endl;
  return 0;
}

-- 
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www.
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 users: <http://www.cpax.org.uk/prg/writings/googly.php>
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