on Thu Sep 20 2007, Gennadiy Rozental wrote:

Hi,

In my new company we are using Boost.Python. Boost version is 1.33.1.

It's my first steps with the library, so please bare with me if this is something trivial.

I've got this noncopyable class Foo

class Foo : boost::noncopyable { /*...*/ };

and shift operator defined:

Foo& operator<<( Foo& foo, int ) { ... return foo; }

Here is my export statement (FooSP is smart pointer on Foo

using namespace boost::python;

class_( "Foo", no_init ) .def( self << int() ) ;

Here is my Python script extract that is being executed from C++

"foo << 2 << 4"

I never got to the second shift call. Instead I an getting exception from Boost.Python:

TypeError: No to_python (by-value) converter found for C++ type: class A

Where does class "A" come from? Surely it says "class Foo?"

I kinda found couple ways to deal with it:

1. define operator<< to return FooSP, which I do not like for assymetry and other reasons 2. avoid using self << int() and define explicit function: Foo& shift_op( Foo& foo, int ) { ... return foo; }

... .def( "__lshift__", &shift_op, return_internal_reference<1>() )

What is the proper way to deal with it? And why original error is reported?

You'd need to add call policies to your << as you do for __lshift__. Foo is noncopyable, so it can't be converted (by value) to Python. For safety reasons Boost.Python assumes that if a reference is returned from a wrapped function, the referent should be _copied_ into the resulting Python object. Unfortunately, we don't give you a way to add call policies when using the 'self' technique, so option #2 seems to be your best workaround. -- Dave Abrahams Boost Consulting http://www.boost-consulting.com The Astoria Seminar ==> http://www.astoriaseminar.com