Why does "g++ -pedantic" complain about the const_iterator typedef
below? Yet the alternative does compile (when BROKEN is defined).

The error message I get from gcc-2.95.1 is:
----
template.cc: In function `void getMenu(int (*)(Y, X *))':
template.cc:14: warning: ANSI C++ forbids typedef which does not specify
a type
template.cc:14: typedef name may not be class-qualified
template.cc:14: parse error before `;'
----
(basically is it a compiler bug or am I doing something wrong?)

#include <map>

typedef struct ppkg * PPKG;
int get_id(PPKG, int *);

template <class X, class Y>
void getMenu(int (*)(Y, X*))
{
#ifndef BROKEN
  // g++ -pedantic fails
  typedef map <X, Y> DescMap;
#else
  typedef map <int, PPKG> DescMap;
#endif
  typedef DescMap::const_iterator DescIterator;
}

void ppkg_member()
{
  getMenu(get_id);
}
---
[ comp.std.c++ is moderated.  To submit articles, try just posting with ]
[ your news-reader.  If that fails, use mailto:std-c++@ncar.ucar.edu    ]
[              --- Please see the FAQ before posting. ---               ]
[ FAQ: http://reality.sgi.com/austern_mti/std-c++/faq.html              ]

In article <1sr02u$2b8@info2.rus.uni-stuttgart.de> kocher@nvdv.e-technik.uni-stuttgart.dbp.de (Hartmut Kocher) writes:
>In article <1993May12.130226.21457@cs.kuleuven.ac.be>, herman@cs.kuleuven.ac.be (Herman Moons) writes:
>> I've got a problem with templates.
>stuff deleted
>>      |       void f( B<A<T3>> );    // syntax error
>
>Standard problem - standard solution:
>void f( B<A<T3> > ); will do the trick. If you don't include the blank,
>'>>' is parsed as the right shift operator.
>
>Hope this helps
>Hartmut

The changes to the lexer and parser to make this syntax work are remarkably
simple.  The lexer needs to return the token >> as a pair of tokens, the
first being a token "> followed immediately followed by another one" and the
second being >.  The expression grammar needs to be modified to accept the
new form of right shift.  A special top-level expression grammar that does not
allow the > or >> operators is needed for templates.  Within that top-level
grammar, parenthesized expressions should use the normal expression grammar.

When I last checked, the syntax of templates was such that allowing
expressions containing the > and >> operators in template declarations did not
lead to syntactic ambiguities.  Nonetheless, an LR(k) grammar for such
declarations is complex and useless for compilation.  I am afraid that
allowing this generalization would require the use of yet another oracle.

     Yours truly,
     Robert Corbett

I've got a problem with templates. I want to re-use some template classes
but C++ (or the AT&T compiler??) won't let me.
The code fragment below illustrates the problem.


     |  template <class T1>
     |  class A {
     |       // ...
     |  };
     |
     |  template <class T2>
     |  class B {
     |       // ...
     |   };
     |
     |  template <class T3>
     |  class C {
     |       // ...
     |       void f( B<A<T3>> );    // syntax error
     |  };
     |
     |  main () { C<int> c; }

Why the syntax error ? Is there a way around this problem ? Any suggestions
are greatly appreciated.
--
Herman Moons                                   Katholieke Universiteit Leuven
e-mail: herman@cs.kuleuven.ac.be               Dept. of Computer Science
                                               Celestijnenlaan 200A
                                               B-3001 Heverlee-Leuven

In article <1993May12.130226.21457@cs.kuleuven.ac.be>, herman@cs.kuleuven.ac.be (Herman Moons) writes:
> I've got a problem with templates.
stuff deleted
>      |       void f( B<A<T3>> );    // syntax error

Standard problem - standard solution:
void f( B<A<T3> > ); will do the trick. If you don't include the blank,
'>>' is parsed as the right shift operator.

Hope this helps
Hartmut