[C++-sig] Re: How to specify a double return type?

[Ralf W. Grosse-Kunstleve]

--- brett hartshorn <bhartsho at yahoo.com> wrote:
> Thanks Ralf and Nicolas for answering my first question.  boost::noncopyable
> fixed my problem!
> 
> Sadly, dReal is a typedef.  I have tried to make 'thin wrappers' but i am not
> sure if i'm doing it
> right.  Here's an example, Foo is the class i am wrapping, Bar is my thin
> wrapper.
> 
> typedef double dReal;
> 
> class Foo {
>   public:
>     const dReal * getSomething() const {
>       return otherFunction();
>     }
> 
> };
> class Bar : public Foo {
>   public:
>     const double * getSomething() const {
>        return Foo::getSomething();
>     }
> };

You are making a "thick wrapper". And as far as Boost.Python is concerned there
is no difference between wrapping Foo::getSomething and Bar::getsomething.

I believe you have to do something like this (untested):

// a thin wrapper
double
Foo_getSomething(Foo const& foo_instance)
{
  return *foo_instance.getSomething(); // dereference the pointer
}

class_<Foo>("Foo")
  .def("getSomething", Foo_getSomething)
;

You can .def an unbound C++ function as a Python method. Cool?

In Python foo_instance.getSomething() returns the *current* value of your
dReal. If the value inside your Foo class changes you have to call
foo_instance.getSomething() again to see the change in Python. E.g. instead of
holding on to the const double* (as you might in C++) you have to hold on to
the foo_instance instead.

HTH,
        Ralf


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