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ttanner2@bloomberg.net
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 Posted: Thu Dec 23, 2004 11:47 am Post subject: Confusion with "friend" declarations |
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Some things which are currently confusing me with friend declarations -
gcc (3.5.0 20040801) gives error messages, and sun workshop 8 doesn't.
Mind you, what the sun compiler lets through as valid is pretty
worrying. Anyway.
a)
class wurble { protected: int somefunc(); };
class burble { friend int wurble::somefunc() };
won't compile on gcc 'int wurble::somefunc() is protected'. Is this a
requirement of the language, or is it a problem with gcc. If for
instance, I just make the whole of wurble a friend of burble this error
doesn't happen, which seems just silly.
b)
class wurble;
class burble { friend int wurble::somefunc() };
won't compile either, because wurble::somefunc() isn't defined.
c) Completely different problem
#include <iostream>
template <size_ N, size_t M> class Wibble
{
friend std::ostream& operator<< <> (std::ostream &, Wibble
const &)
};
refuses to compile unless I predeclare operator<<. It's OK if the
template only has one parameter. Is this an issue with gcc or is it a
language thing.
Thanks
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Alberto Barbati
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| Posted: Thu Dec 23, 2004 9:48 pm Post subject: Re: Confusion with "friend" declarations |
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[email]ttanner2 (AT) bloomberg (DOT) net[/email] wrote:
| Quote: | Some things which are currently confusing me with friend declarations -
gcc (3.5.0 20040801) gives error messages, and sun workshop 8 doesn't.
Mind you, what the sun compiler lets through as valid is pretty
worrying. Anyway.
a)
class wurble { protected: int somefunc(); };
class burble { friend int wurble::somefunc() };
won't compile on gcc 'int wurble::somefunc() is protected'. Is this a
requirement of the language, or is it a problem with gcc.
|
gcc is correct. §11.4/7 states that "A name nominated by a friend
declaration shall be accessible in the scope of the class containing the
friend declaration." In this case wurble::somefunc is not accessible in
the scope of class burble.
| Quote: | instance, I just make the whole of wurble a friend of burble this error
doesn't happen, which seems just silly.
|
Doesn't seem so silly to me... why do you say so?
| Quote: | b)
class wurble;
class burble { friend int wurble::somefunc() };
won't compile either, because wurble::somefunc() isn't defined.
|
gcc is correct, because you can't refer to members of an incomplete
class in any context (not only friend declarations).
| Quote: | c) Completely different problem
#include <iostream
template
{
friend std::ostream& operator<< <> (std::ostream &, Wibble
const &)
};
refuses to compile unless I predeclare operator<<.
|
gcc is correct. You can't refer to a particular template specialization
until the generic template has been declared (in any context, not only
friend declations).
| Quote: | It's OK if the template only has one parameter.
|
What do you mean by that? Could you please elaborate?
Alberto
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BigBrian
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| Posted: Thu Dec 23, 2004 10:12 pm Post subject: Re: Confusion with "friend" declarations |
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| Quote: | a)
class wurble { protected: int somefunc(); };
class burble { friend int wurble::somefunc() };
won't compile on gcc 'int wurble::somefunc() is protected'. Is this a
requirement of the language, or is it a problem with gcc.
|
You've put the "friend" keyword in the wrong class. If you want burble
to access wurble you need the following
class wurble
{
friend class burble;
protected:
int soomefunc();
};
class burble
{
void foo() { wurble w; w.somefunc(); }
};
-Brian
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ThosRTanner
Guest
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| Posted: Fri Dec 24, 2004 9:47 am Post subject: Re: Confusion with "friend" declarations |
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I'll try to answer this comprehnsibly - google groups beta doesn't
appear to do quotuing. Oh well, here goes:
| Quote: | a)
class wurble { protected: int somefunc(); };
class burble { friend int wurble::somefunc() };
won't compile on gcc 'int wurble::somefunc() is protected'. Is this
a
requirement of the language, or is it a problem with gcc.
gcc is correct. §11.4/7 states that "A name nominated by a friend
declaration shall be accessible in the scope of the class containing
the
friend declaration." In this case wurble::somefunc is not accessible
in
the scope of class burble.
instance, I just make the whole of wurble a friend of burble this
error
doesn't happen, which seems just silly.
Doesn't seem so silly to me... why do you say so?
|
Because
a) Wurble::somefunc is NOT USED by Burble. Why should it's access class
affect whether or not it can be a friend?
b) Instead of saying only one particular function is a friend of my
class, and therefore allowed to access it, I'm saying the whole of a
particular class is a friend of my class, and therefore anything in
that class is allowed to access it.
c) By declaring the whole class a friend, I'm allowing protected and
private functions in that class to access my class. But I cannot
nominate individual functions as friends. This doesn't strike me as
entirely sensible.
It would appear that 11.4/7 is forcing me to reduce the safeness of my
class.
| Quote: | c) Completely different problem
#include <iostream
template
{
friend std::ostream& operator<< <> (std::ostream &, Wibble
const &)
};
refuses to compile unless I predeclare operator<<.
gcc is correct. You can't refer to a particular template
specialization
until the generic template has been declared (in any context, not
only
friend declations).
It seems a very roundabout way of doing things. I realise (now) that a |
friend declaration doesn't constitute a definition in scope, but it is
really aggravating to have to write 2 lines of declaration to declare
something as a friend later on. What is the rationale behind that
decision?
| Quote: | It's OK if the template only has one parameter.
What do you mean by that? Could you please elaborate?
As in |
#include
template <size_ N> class Wibble
{
friend std::ostream& operator<< <> (std::ostream &, Wibble const &)
};
seems to work.
I think it might be a featurette of gcc - I have discovered that the
above works sometimes and not others - it appears to be connected with
other declarations of operator<< having been found.
[ See http://www.gotw.ca/resources/clcm.htm for info about ]
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Alberto Barbati
Guest
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| Posted: Tue Dec 28, 2004 10:38 am Post subject: Re: Confusion with "friend" declarations |
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ThosRTanner wrote:
| Quote: |
c) Completely different problem
#include <iostream
template
{
friend std::ostream& operator<< <> (std::ostream &, Wibble
const &)
};
refuses to compile unless I predeclare operator<<.
gcc is correct. You can't refer to a particular template specialization
until the generic template has been declared (in any context, not only
friend declations).
It seems a very roundabout way of doing things. I realise (now) that a
friend declaration doesn't constitute a definition in scope, but it is
really aggravating to have to write 2 lines of declaration to declare
something as a friend later on. What is the rationale behind that
decision?
|
I guess because the compiler has to determine the complete signature of
the friend function immediately at the point of the friend declaration.
For a function template specialization, the signature includes the
"values" of all template parameters. Remember that the line:
friend std::ostream& operator<< <>
(std::ostream &, Wibble const &);
really should be (and is interpreted as):
friend std::ostream& operator<<
(std::ostream &, Wibble<N> const &);
The second "N" is implicit because "Wibble" is an alias for "Wibble<N>"
in that context, but the first "N" can only be deduced by comparing the
function parameters with the template declaration. If the compiler
hasn't already seen the template declaration, how could it determine
that you are not referring to, for example:
friend std::ostream& operator<<
(std::ostream &, Wibble<N> const &);
or even something else completely different?
You may find it complicated and I might even agree with you, but put
yourself in the compiler's shoes...
| Quote: | It's OK if the template only has one parameter.
What do you mean by that? Could you please elaborate?
As in
#include <iostream
template
{
friend std::ostream& operator<< <> (std::ostream &, Wibble const &)
};
seems to work.
|
It shouldn't and it doesn't on both VC++ 7.1 and Comeu Online, they both
don't accept the code as it is. Gcc 3.3.3 accepts the code, but produces
an error as soon as try to instantiate the Wibble template.
| Quote: | I think it might be a featurette of gcc - I have discovered that the
above works sometimes and not others - it appears to be connected with
other declarations of operator<< having been found.
|
The presence of other declarations of operator<< might affect the
parsing and confuse the compiler, but unless you declare a matching
template before the friend declaration, the code will still be ill-formed.
Alberto
[ See http://www.gotw.ca/resources/clcm.htm for info about ]
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Alberto Barbati
Guest
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| Posted: Tue Dec 28, 2004 10:34 pm Post subject: Re: Confusion with "friend" declarations |
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