FAQ: substituting cout (C++)

As I know, cout is object of an ostream class. Am I right?

I have the following question: I would like to use another 'command' to output data on the screen.
e.g. instead of

cout << "abc";

i would like to use

my_own_cout << "abc";


I have tried to declare it:
ostream my_own_cout;
But it didn't work. Why is that so?

How to make a 'substitute' for cout?

TNX

I think macro's are what you're looking for,
Don't know exactly how to use it with 'cout' though

#define cout my_own_cheese

Quote:

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Originally posted by Shadow12345
#define cout my_own_cheese

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You mixed up the order, it would be

#define my_own_cheese cout

Still, Sang-drax is right, #defines are evil!

No, #defines are evil

Code:

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std::ostream& my_own__cout = cout;

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Thanx sang-drax, the code is working. But can someone explain it to me a little bit?

Code:

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std::ostream& my_own_cout = cout;

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why it is not enough to write:

Code:

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ostream my_own_cout;

or

std::ostream my_own_cout;

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if cout is an object of class ostream?

Thank you in advance!

>std::ostream& my_own_cout = cout;

Is creating a reference (alias) to the original cout. my_own_cout isn't a seperate object. If you really want a seperate ostream object you could do -

ostream my_own_cout(cout.rdbuf());

Which will point your ostream streambuf pointer at the same one cout uses. Or if that's not enough you could derive you're own streambuf class an implement all the system calls, etc and assign it to your ostream object.

cout is the standard output device, so you can access it directly, under a different name by using a reference. To declare an ostream object you have to pass in a streambuf pointer, I don't think it has a default constructor.

EDIT: as was already said... hehe =)