🚀 go-pugleaf

tim@franck.Princeton.EDU (Tim Hollebeek) wrote:

> I suspect this is correct behavior, but it sure is confusing, so I
> thought I'd double check:
>
> template <class T>
> class nVector {
> };
>
> typedef nVector<double> Vector; // workaround for compiler that
>                                 // doesn't handle <class T = double> > 
>
> class Point : public Vector { // derive from typedef name
> public:
>    Point(const Vector& v) { Vector::Vector(v); } // ***
> };
>
> At ***, I get: test.C:9: no method `nVector<double>::Vector'
> for obvious reasons.  However, it's not so obvious if you are just
> looking at the declaration of class Point, and the alternatives:

The reason that 'Vector::Vector(v)' doesn't work is not that Vector is a 
typedef.  In fact for most (all?) purposes, the typedef name is as good 
as a class.  The problem is that the way to explicitly call a 
constructor is by using the name of the constructor *unqualified* by the 
scope of the class.  It's as though the constructor were defined in 
namespace enclosing the class, rather than that of the class itself.  So 
for a normal member function, you call it by qualifying it by the class 
name (Vector::insert), but for a constructor, you refer to it just by 
its name (Vector).  Your example would work with:
    Point (const Vector& v) { Vector(v); }
although this will just create a temporary variable, initialise it, and 
then discard it (as your original code would do too, if it worked).
 
> Vector::nVector<double>(v); // parse error
> nVector<double>::nVector<double>(v); // parse error

Same applies.  Simply 'nVector<double>(v);' would be sufficient.  Again, 
such code is pointless unless nVector<double>'s copy constructor has 
side effects.
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