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Roger
Guest
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 Posted: Tue Aug 29, 2006 3:06 am Post subject: Calling virtual method of base class nested in template clas |
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Hi all.
Here's a little quandry I've run in to and can't tell if it's a compiler
bug, or if it's bad code. First, here's the code:
===================================
#include <stdio.h>
template < typename T >
class B
{
public:
B( ) { }
class N1
{
public:
N1( ) { };
class N2
{
public:
N2( ) { };
virtual void f( )
{
printf( " B< >::N1::N2::f( )\n" );
}
};
};
T _m;
};
class D : public B< int >
{
public:
class N1 : public B< int >::N1
{
public:
N1( ) : B< int >::N1( ) { };
class N2 : public B< int >::N1::N2
{
public:
N2( ) : B< int >::N1::N2( ) { };
virtual void f( )
{
printf( " D::N1::N2::f( )\n" );
( ( B< int >::N1::N2* ) this )->f( );
}
};
};
};
void
main( int argc, char* argv[ ] )
{
D::N1::N2 n2;
n2.f( );
}
==================================
Invoking n2.f( ) causes an infinite loop, dumping out "D::N1::N2::f( )"
ad-nausium. I am trying to call the base-class f( ) in B::N1::N2, but the
compiler just generates code that instead calls D::N1::N2::f( ) recursively.
What am I doning wrong? How can I call the base's f( )?
Thanks for any help you can offer!
- Roger
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Jens Theisen
Guest
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| Posted: Tue Aug 29, 2006 4:05 am Post subject: Re: Calling virtual method of base class nested in template |
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Roger wrote:
| Quote: | What am I doning wrong? How can I call the base's f( )?
|
This way:
virtual void f( )
{
printf( " D::N1::N2::f( )\n" );
B< int >::N1::N2::f( );
}
:)
Jens
[ See http://www.gotw.ca/resources/clcm.htm for info about ]
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Roger
Guest
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| Posted: Tue Aug 29, 2006 8:26 pm Post subject: Re: Calling virtual method of base class nested in template |
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|
{ please watch your quoting. did you actually need to quote? -mod }
Actually, I came up with a simpler example that exhibits the same bug (no
template, and nesting at only a single level):
#include <iostream>
class B
{
public:
B( ) { }
class N1
{
public:
N1( ) { };
virtual void f( )
{
std::cout << "B< >::N1::f( )" << std::endl;
}
};
};
class D : public B
{
public:
class N1 : public B::N1
{
public:
N1( ) : B::N1( ) { };
virtual void f( )
{
typedef B::N1 BN1;
std::cout << "D::N1::f( )" << std::endl;
BN1::f( );
}
};
};
void
main( int argc, char* argv[ ] )
{
D::N1 n1;
n1.f( );
}
However, the problem is worked around by changing the name of D::N1 to
something other than N1. Apperently, the VC6.0 compiler becomes confused as
to which N1 I was referring and just assumed the most derived N1, rather
than the one specified by the typedef (i.e., the N1 in B).
When I renamed N1 to, say, Joe, it worked.
Thanks,
- Roger
"Jens Theisen" <jth02 (AT) arcor (DOT) de> wrote in message
news:44f36da0$0$20038$9b4e6d93 (AT) newsspool4 (DOT) arcor-online.net...
| Quote: | Roger wrote:
What am I doning wrong? How can I call the base's f( )?
This way:
virtual void f( )
{
printf( " D::N1::N2::f( )\n" );
B< int >::N1::N2::f( );
}
:)
Jens
|
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kanze
Guest
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| Posted: Tue Aug 29, 2006 11:52 pm Post subject: Re: Calling virtual method of base class nested in template |
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|
Roger wrote:
| Quote: | Here's a little quandry I've run in to and can't tell if it's
a compiler bug, or if it's bad code. First, here's the code:
|
Which you really could have considerably simplified. The
problem has nothing to do with nested classes.
| Quote: | ===================================
#include <stdio.h
template < typename T
class B
{
public:
B( ) { }
class N1
{
public:
N1( ) { };
class N2
{
public:
N2( ) { };
virtual void f( )
{
printf( " B< >::N1::N2::f( )\n" );
}
};
};
T _m;
};
class D : public B< int
{
public:
class N1 : public B< int >::N1
{
public:
N1( ) : B< int >::N1( ) { };
class N2 : public B< int >::N1::N2
{
public:
N2( ) : B< int >::N1::N2( ) { };
virtual void f( )
{
printf( " D::N1::N2::f( )\n" );
( ( B< int >::N1::N2* ) this )->f( );
|
Which invokes the virtual function mechanism (and probably
results in a endless recursion).
How should this be any different from:
B< int >::N1::N2* p ;
p->f() ;
invoked on any object, anywhere.
If you want to invoke the base class function, you have to say
so:
p->B< int >::N1::N2::f() ;
(or in this case, the same thing, without the p->).
You mean 'int', of course. void is not a legal return value for
main.
| Quote: | main( int argc, char* argv[ ] )
{
D::N1::N2 n2;
n2.f( );
}
==================================
Invoking n2.f( ) causes an infinite loop, dumping out
"D::N1::N2::f( )" ad-nausium.
|
It's not looping, it's recursing. And it's only conceptually
infinite: you'll run out of stack sooner or later, and then it
will probably stop.
| Quote: | I am trying to call the base-class f( ) in B::N1::N2, but the
compiler just generates code that instead calls D::N1::N2::f()
recursively.
|
Typical case of the compiler doing what you said, and not what
you wanted.
| Quote: | What am I doning wrong? How can I call the base's f()?
|
By using a qualified name for the function.
--
James Kanze GABI Software
Conseils en informatique orientée objet/
Beratung in objektorientierter Datenverarbeitung
9 place Sémard, 78210 St.-Cyr-l'École, France, +33 (0)1 30 23 00 34
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