Re: [Boost-Users] lambda library bug --- endl?

From: Kenneth Yarnall (yarnall_at_[hidden])
Date: 2002-09-13 12:22:18

On Fri, 2002-09-13 at 12:22, Jaakko Jarvi wrote:
> Hi Ken,
> 
> >     for_each(ell.begin(), ell.end(), cout << _1 << endl);
> 
> > Any suggestions?  I am aware that I can replace endl by '\n' and get a
> > compilable program.  I worry that this problem extends to other
> > manipulators...
> > 
> 
> It's not a bug, just an unfortunate restriction.

Thanks for the quick response.

> Example: endl is not now allowed in std::cout statements in a Lambda 
> expression. '\n' has to be used instead.
> 
> -- This is because the old style streams were not templates and the new 
> style ones are. Also, endl has been changed from a non-template function 
> to a template function and one cannot take the address of a function 
> template (withtout instantiated it at the same time. For example, in _1 << 
> endl, the library does not know what the type of the stream that later is 
> substituted for _1 will be. -- Jaakko Jarvi 

So if I explicitly list the template parameters, does it work?  Hmmm...

No, this doesn't work:

for_each(ell.begin(), ell.end(), 
    cout << _1 << endl<char, char_traits<char> >);

I was hopeful --- I'm pretty sure I have the syntax right.  I'm simply
curious now: is it possible to make this work?

Ken
-- 
Ken Yarnall
Dept. of Mathematical Sciences               Lebanon Valley College
Assoc. Professor of Math and CS       [phone] yarnall_at_[hidden]