Reputation: 1061
game.h
#ifndef GAME_H
#define GAME_H
#include <string>
#include <iostream>
#include "piece.h"
using namespace std;
class Game
{
private:
string white;
string black;
string title;
istream* in;
ostream* out;
public:
Game();
Game(istream&, ostream&);
void display(Colour, short);
};
#endif
game.cpp
#include <iostream>
#include <string>
#include <sstream>
#include "game.h"
#include "board.h"
#include "piece.h"
using namespace std;
Game::Game()
{
//nothing
}
Game::Game(istream& is, ostream& os)
{
in = is;
out = os;
}
void Game::display(Colour colour, short moves)
{
//out << "a";
}
I'm trying to use the istream and ostream in other parts of my class but I can't because g++ won't let me reference is to in. Any ideas?
Upvotes: 7
Views: 10054
Reputation: 1151
You should deference the pointer:
*out << "a";
For more convenient use, to not deference the pointers each time, and for more readability you can use references instead of pointers.
class Game
{
// ...
std::istream& in; // notice explicit namespace std::
std::ostream& out;
// ...
};
Then you can write:
out << "a";
Plus, it is not a good habit to do so:
using namespace std;
This way you are exposing the names of std namespace.
Upvotes: 4
Reputation: 137810
You simply want a reference variable, not a pointer.
class Game
{
private:
...
istream& in;
ostream& out;
public:
Game(istream&, ostream&);
};
Game::Game(istream& is, ostream& os)
: in( is ),
out( os )
{ }
The existing code compiles because of a couple language quirks:
istream
/ ostream
are convrtible to void*
to allow you to check their error status as in
if( in ) { do_something( in ); }
your compiler apparently allowed void*
to be converted to ostream*
(I believe in error, and you should at least get a warning from this).
Upvotes: 7
Reputation: 92261
If you store pointers, you need to dereference them, like *in
or *out << ...
.
Upvotes: 1
Reputation: 36433
is
is a reference not a pointer, therefore if you want to store a pointer you need to use the address operator in = &is;
But please realize that is
can cease to exist immediately after the method call, therefore you can easily end up with an invalid pointer. Make sure that you at least document this fact.
Upvotes: 2