Consider a class template like the following, that when instantiated
will probably generate some warnings (I know that warnings are outside
the standard, but anyway):

template< int N > struct Deprecated
  {
  Deprecated( int ) {}
  ~Deprecated()
    {
    int* deprecated;
    *deprecated = 0 ;  // vc++, bc++ catch this
    deprecated[3] ;    // gcc 2.95, -W[all] catch this
    }
  } ;


Now consider a function, taking a reference to a specialization of that
class as a parameter, with a default value that will cause a temporary
to be created:

// prototype
void f( int& a , const Deprecated< 0 >& d = 0 ) ;

// definition
void f( int& a , const Deprecated< 0 >& )
  {
  ++a ;
  }

Obviously, if you include all the above in a compilation unit, followed
by a main() like this:

int main()
  {
  int x = 3 ;
  f( x ) ;         // line XXX
  return 0 ;
  }

you get a warning, which (with some luck) you can trace back to line
XXX.

My question is: what should happen if you comment out line XXX? A few
compilers that I have tried (Visual C++ 6, C++ Builder 5, gcc 2.95),
seeing no *calls* to f, will not instantiate Deprecated, and no warning
will be generated. Still, with other compilers (like Comeau 4.2.43) you
always get a warning.

The real issue is: does the declaration/definition of f cause an
implicit instantiation of Deprecated< 0 >? From my understanding of the
standard it should not: 14.7.1p9 says "the use of a template
specialization in a default argument shall not cause the template to be
implicitly instantiated except that a class template may be
instantiated where its complete type is needed to determine the
correctness of the default argument". The "except ..." part
does not seem to apply (?), so I would expect Deprecated to be
instantiated, and a warning to be generated, *only* if f() is actually
called.

Opinions, suggestions, reports from other compilers would be much
appreciated.

Have fun,
Carlo Pescio



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