MESSAGE
| DATE | 2011-04-12 |
| FROM | Ruben Safir
|
| SUBJECT | Subject: [NYLXS - HANGOUT] [billy.donahue@gmail.com: Re: templated operator()<<]
|
For tonights discussion
Tuesday night NYLXS C++ workshop - 7:00PM
1163 east 15th street
Brooklyn, USA 11230
Q or B to Avenue J
----- Forwarded message from Billy -----
Date: Tue, 12 Apr 2011 14:34:17 -0400
From: Billy
To: Ruben Safir
Subject: Re: templated operator()<<
ohhhhhhhhhh! you were already inside a template class! Didn't realize
that.
This is why it's important to give the full situation.
template
friend std::ostream& operator<<(std::ostream&, const Distribution&);
Yes, this is what you do, yes, fine.
template
std::ostream& operator<<(std::ostream& os, const Distribution& obj) {
T desc = obj.description();
int pop = obj.population();
os << "The Identification of " << desc << " was seen " << pop ;
return os;
}
..Okay, so here's the deal about this:
86 template friend std::ostream& operator<<(std::ostream&, const
Distribution&);
test_del.cpp:86: error: declaration of ?class T?
test_del.cpp:44: error: shadows template parm ?class T?
You're accidentally declaring a new function base template here, taking a
class T as parameter.
The symbol 'T' is already used as a parameter of the enclosing class
template class Distribution {...};
SO when you use 'T' in the argument list later (the 'const Distribution&'
argument), it's warning you
that you'll be using a different 'T', one that's local to the friend
definition.
Here's an incorrect fix, which might be useful to think about:
86 template friend std::ostream& operator<<(std::ostream&, const
Distribution&);
This grants friendship from all overloads of operator<<(std::ostream&,const
Distribution&) for any U to all Distribution classes.
That is to say, operator<<(std::ostream&, const Distribution&) can, in
its body, look at private members of the class
Distribution, which is not what you want.
What YOU want, is for a single specialized template function, called:
template std::ostream& operator<<(std::ostream&, const
Distribution&)
to have access to the privates of the Distribution. This is not a family
of functions, this is one function.
Anyway, since you're using only public accessors in the body of operator<<,
you can get by without the friend
line altogether. Just take it out.
On Tue, Apr 12, 2011 at 2:06 PM, Ruben Safir wrote:
> The only one that is needed to compile is test_del.cpp
>
> Ruben
> --
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--
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----- End forwarded message -----
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