I am fairly new to object oriented programming, but have two simple questions on a piece of code. Note that this code has been simplified greatly without changing the net result:

#include <stdio.h>
#include <iostream.h>

class first_one
{
int x;
int y;
public: first_one()
{
  x = 1;
  y = 2;
}
};

class second_one
{
  int z;
  public: second_one()
{
  z = 3;
}

void print_it()
{
  printf("%d %d %d %d %d %d %d");
}
};

void main()
{
  first_one A;
  second_one B;
  B.print_it();
}

The following is printed:

-403 592 -221 3 2 1

Why does this occur?
Also, what C++ or object oriented principle did I violate?

Re: Why do I get the following printf() result? 80 80

Well you wanted C++ but you sort of mashed together non-OO C with OO C++. Also the .h antiquates the header and there is no real reason to have stdio.h in there since that is a C header. But if you are going to use printf the way you used it I'm surprised even compiles since you don't actually give printf any variables.

int x = 10;
printf("This is an integer: %i", x);

would print an integer

You should really just use cout if you are using C++

int x = 10;
cout<<"This is an integer: "<< x << endl;

is the C++ equivalent

Re: Why do I get the following printf() result? 80 80

>>I'm surprised even compiles
Depends on the compiler. Some compilers check the first argument to see if any other arguments are needed and give warnings for mismatches. Other compilers don't care.

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