[lug] simple iostream bug?

[Jonathan Briggs]

A quick correction to myself.  I was curious about the endl manipulator 
thing and so I looked it up.  The endl function is declared as ostream& 
endl(ostream&).  Basically. (Lots of template stuff in there with it.) 
 When used just as endl, it is a function pointer.  The iostreams 
library has a declaration to deal with these by calling the function 
pointer on the stream.

So your TextHeader function should work fine if you just call it as cout 
<< TextHeader << endl;

Jonathan Briggs wrote:

> This is because you are sending a ostream reference (the return value 
> of TextHeader) into an ostream and it appears to be converting it into 
> an unsigned long.  This is probably because iostreams return true if 
> the stream state is good and false if the stream state is bad (so you 
> can do while( cout ), or similar things.)
>
> Just do TextHeader( cout ).
>
> If you want to be able to do cout << TextHeader (as a manipulator, 
> like how endl is used) you'll need to do some more magic (Bear with 
> me, I'm not sure I remember this perfectly.  You may need to consult 
> documentation.):
>
> Create a manipulator class/struct:  This can be anything, it doesn't 
> have to be part of any hierarchy.
> If it takes an argument, you make the constructor take the arguments. 
> If it doesn't take arguments, I think you need to create a global 
> instance of your class (I _think_ endl is an instance of something.)
> Make a non-member function for ostream& operator<<(ostream&, 
> [your_manipulator_class])

[snip]