[C++-sig] Suppressing assignment operator in Boost.Python

[Roman Yakovenko]

On 4/21/06, Abhi <abhi at qualcomm.com> wrote:
> namespace bp = boost::python;
>
>
> BOOST_PYTHON_MODULE(Boost_YMI){
>   bp::class_< B, boost::noncopyable>( "B" )
>         .def( bp::init< >()[bp::default_call_policies()] )
>        .def_readwrite("aObj", &B::aObj)
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

This one has been fixed and unit test has been added.
Now pyplusplus will generate def_readonly.
You will not be able to assign new object to aObj, but you will be
able to modify
it's state.

Quick work around is next:
mb = module_builder_t( .... )
cls_b = mb.class_( 'B' )
cls_b.member_variable( 'aObj' ).exclude()
cls_b.add_code( 'def_readonly( "aObj", &B::aObj )' )

>     ;
>
>
>     bp::class_< A, boost::noncopyable >( "A" )
>         .def( bp::init< >()[bp::default_call_policies()] )
>         .def_readwrite( "x", &A::x );
> }
>
>
> But this does not compile since Boost.Python generates an assignment
> operator for the attribute aObj (which has a private assignment operator).
>
> Is this anyway to tell Boost.Python to not generate the assignment operator
> for aObj of B.

I am not sure that I fully understand you, but proposed solution
should work for you.

> One workaround (which is not at all elegant!) is for me to write the
> following hand-crafted methods:
>
> void set(B& self, A& a)
> {
>   self.aObj.x = a.x;
> }
>
> const A& get(B& self)
> {
>   return self.aObj;
> }
>
>  and then not expose the aObj as an attribute but rather through these
> methods in python:
>     bp::class_< B, boost::noncopyable>( "B" )
>         .def( bp::init< >()[bp::default_call_policies()] )
>         .def("set", &set)
>         .def("get", &get, bp::return_internal_reference<>() )

I don't think you need this.

>
> thanks
> - Abhi


--
Roman Yakovenko
C++ Python language binding
http://www.language-binding.net/