Hi,
I have written custom converters from Python to C++ types and vice versa.
Now I'd like to wrap C++ functions that expect/return shared pointers to
those types.
I could of course write another set of converters for each type, but that
would mean duplicating a lot of code for no good reason. Is there a simple
way to generate converters for shared_ptr<T>, if converters for T already
exist?
See below for a simple test case. From Python I can call test.foo(42) and
get back the same value. Calling test.bar(42) fails, because there's no
converter from int to shared_ptr<Test>. What's the easiest way to make
bar() behave the same way as foo()?
Thanks in advance,
Dominic
--- test module ---
#include <boost/python.hpp>
#include <boost/shared_ptr.hpp>
namespace bp = boost::python;
struct Test {
int value;
};
struct test_from_python_converter {
test_from_python_converter() {
bp::converter::registry::push_back(
&convertible, &construct, bp::type_id<Test>());
}
static void *convertible(PyObject *obj) {
return PyInt_Check(obj) ? obj : 0;
}
static void construct(PyObject *obj,
bp::converter::rvalue_from_python_stage1_data *data) {
void *storage = (reinterpret_cast<bp::converter::
rvalue_from_python_storage<Test>*>(data))->storage.bytes;
new (storage) Test();
((Test *)storage)->value = bp::extract<int>(obj);
data->convertible = storage;
}
};
struct test_to_python_converter
: bp::to_python_converter<Test, test_to_python_converter, true>
{
static PyObject *convert(Test const & t) {
return PyInt_FromLong(t.value);
}
static PyTypeObject const *get_pytype() {
return &PyInt_Type;
}
};
Test foo(Test t) {
return t;
}
boost::shared_ptr<Test> bar(boost::shared_ptr<Test> pt) {
return pt;
}
BOOST_PYTHON_MODULE(test) {
test_from_python_converter();
test_to_python_converter();
bp::def("foo", &foo);
bp::def("bar", &bar);
}
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