Boost.Python: Defining a constructor outside a class

boost-python, constructor, factory

Solution

You can use `make_constructor` (untested):

TCurrency* TCurrency_from_Foo( const Foo& ) { return new TCurrency(); }

class_<TCurrency>( "TCurrency" )
    .def( "__init__", boost::python::make_constructor( &TCurrency_from_Foo) )
;

The argument to make_constructor is any functor that returns a pointer[1] to the wrapped class.

[1] Actually, the function must return a the pointer holder type, so if your pointer holder is `boost::shared_ptr`, the function should return a boost::shared_ptr instead of a raw pointer.

Problem

Given a class: ``` class TCurrency { TCurrency(); TCurrency(long); TCurrency(const std::string); ... }; ``` Wrapped with Boost.Python: ``` class_<TCurrency>( "TCurrency" ) .def( init<long> ) .def( init<const std::string&> ) ... ; ``` Is it possible to create a factory method that appears as a constructor in Python: ``` TCurrency TCurrency_from_Foo( const Foo& ) { return TCurrency(); } ``` Such that in python: ``` bar = TCurrency(foo) ```

Original source