boost python 返回包含自定义类型的元组
boost python returning tuple containing custom types
我有一个 C++ myObject class,我使用包装器结构通过 boost python 公开:
struct myObjectWrapper{
static tuple compute(myObject& o,const Container& x0, const double& t0, Container& x){
double t;
int stat = o.evaluate(x0,t0,x,t);
return make_tuple(stat,t);
}
}
BOOST_PYTHON_MODULE(myModule)
{
// not shown here is code to expose Container class
class_<myObject>("MyObject")
.def("compute",&myObjectWrapper::compute)
;
}
容器当前定义为:
typedef std::valarray<double> Container
并暴露于 python。
现在python我可以了。
x = Container()
(status,t) = obj.compute(Container([0.,0.,0.]),0.0,x)
print status, t, x[0]
这不是很 pythonic。我更愿意做:
(status,t,x) = obj.compute(Container([0.,0.,0.]),0.0)
print status, t, x[0]
我可以在 python 中编写一个额外的包装器,但我宁愿避免添加更多包装器。
以下代码无法编译:
struct myObjectWrapper{
static tuple compute(myObject& o,const Container& x0, const double& t0){
double t;
Container x;
int stat = o.evaluate(x0,t0,x,t);
return make_tuple(stat,t,x);
}
}
此外,我更愿意窃取局部变量 x 的内容并让 python 管理它而不是复制它:
return make_tuple(stat,t,std::move(x));
如何实现?
简而言之,在免费存储上分配包装器并使用 manage_new_object
result convert to transfer ownership to a Python object. This will cause Boost.Python to copy the pointer when constructing the Python object, rather than copying the pointee. For more details, see 答案。
这是一个将所有权转移到 Python 对象的辅助函数:
/// @brief Transfer ownership to a Python object. If the transfer fails,
/// then object will be destroyed and an exception is thrown.
template <typename T>
boost::python::object transfer_to_python(T* t)
{
// Transfer ownership to a smart pointer, allowing for proper cleanup
// incase Boost.Python throws.
std::unique_ptr<T> ptr(t);
// Use the manage_new_object generator to transfer ownership to Python.
namespace python = boost::python;
typename python::manage_new_object::apply<T*>::type converter;
// Transfer ownership to the Python handler and release ownership
// from C++.
python::handle<> handle(converter(*ptr));
ptr.release();
return python::object(handle);
}
并且可以如下使用它:
boost::python::tuple myObjectWrapper::compute(
myObject& o, const Container& x0, const double& t0)
{
auto x1 = std::make_unique<container>();
double t1 = 0;
int stat = self.evaluate(x0, t0, *x1, t1);
return boost::python::make_tuple(stat, t1, transfer_to_python(x1.release()));
}
这里是一个完整的例子,基于 demonstrates 使用 transfer_to_python
辅助函数的原始问题。
#include <boost/python.hpp>
#include <cassert>
#include <memory> // std::unique_ptr
// Mock legacy API.
struct container
{
container() {}
container(boost::python::object) {}
container(const container&)
{
// For this example, guarantee copy is not made.
assert(false);
}
};
struct my_object
{
int evaluate(container&, double&, container&, double&) { return 42; }
};
/// @brief Transfer ownership to a Python object. If the transfer fails,
/// then object will be destroyed and an exception is thrown.
template <typename T>
boost::python::object transfer_to_python(T* t)
{
// Transfer ownership to a smart pointer, allowing for proper cleanup
// incase Boost.Python throws.
std::unique_ptr<T> ptr(t);
// Use the manage_new_object generator to transfer ownership to Python.
namespace python = boost::python;
typename python::manage_new_object::apply<T*>::type converter;
// Transfer ownership to the Python handler and release ownership
// from C++.
python::handle<> handle(converter(*ptr));
ptr.release();
return python::object(handle);
}
// API wrapper.
boost::python::tuple my_object_compute(
my_object& self, container& x0, double t0)
{
auto x1 = std::make_unique<container>();
double t1 = 21;
int stat = self.evaluate(x0, t0, *x1, t1);
return boost::python::make_tuple(stat, t1, transfer_to_python(x1.release()));
}
BOOST_PYTHON_MODULE(example)
{
namespace python = boost::python;
python::class_<container>("Container")
.def(python::init<python::object>())
;
python::class_<my_object>("MyObject")
.def("compute", &my_object_compute)
;
}
交互使用:
>>> import example
>>> my_object = example.MyObject()
>>> status, t, x = my_object.compute(example.Container([1, 2, 3]), 4)
>>> assert(status == 42)
>>> assert(t == 21)
>>> assert(isinstance(x, example.Container))
我有一个 C++ myObject class,我使用包装器结构通过 boost python 公开:
struct myObjectWrapper{
static tuple compute(myObject& o,const Container& x0, const double& t0, Container& x){
double t;
int stat = o.evaluate(x0,t0,x,t);
return make_tuple(stat,t);
}
}
BOOST_PYTHON_MODULE(myModule)
{
// not shown here is code to expose Container class
class_<myObject>("MyObject")
.def("compute",&myObjectWrapper::compute)
;
}
容器当前定义为:
typedef std::valarray<double> Container
并暴露于 python。
现在python我可以了。
x = Container()
(status,t) = obj.compute(Container([0.,0.,0.]),0.0,x)
print status, t, x[0]
这不是很 pythonic。我更愿意做:
(status,t,x) = obj.compute(Container([0.,0.,0.]),0.0)
print status, t, x[0]
我可以在 python 中编写一个额外的包装器,但我宁愿避免添加更多包装器。
以下代码无法编译:
struct myObjectWrapper{
static tuple compute(myObject& o,const Container& x0, const double& t0){
double t;
Container x;
int stat = o.evaluate(x0,t0,x,t);
return make_tuple(stat,t,x);
}
}
此外,我更愿意窃取局部变量 x 的内容并让 python 管理它而不是复制它:
return make_tuple(stat,t,std::move(x));
如何实现?
简而言之,在免费存储上分配包装器并使用 manage_new_object
result convert to transfer ownership to a Python object. This will cause Boost.Python to copy the pointer when constructing the Python object, rather than copying the pointee. For more details, see
这是一个将所有权转移到 Python 对象的辅助函数:
/// @brief Transfer ownership to a Python object. If the transfer fails,
/// then object will be destroyed and an exception is thrown.
template <typename T>
boost::python::object transfer_to_python(T* t)
{
// Transfer ownership to a smart pointer, allowing for proper cleanup
// incase Boost.Python throws.
std::unique_ptr<T> ptr(t);
// Use the manage_new_object generator to transfer ownership to Python.
namespace python = boost::python;
typename python::manage_new_object::apply<T*>::type converter;
// Transfer ownership to the Python handler and release ownership
// from C++.
python::handle<> handle(converter(*ptr));
ptr.release();
return python::object(handle);
}
并且可以如下使用它:
boost::python::tuple myObjectWrapper::compute(
myObject& o, const Container& x0, const double& t0)
{
auto x1 = std::make_unique<container>();
double t1 = 0;
int stat = self.evaluate(x0, t0, *x1, t1);
return boost::python::make_tuple(stat, t1, transfer_to_python(x1.release()));
}
这里是一个完整的例子,基于 demonstrates 使用 transfer_to_python
辅助函数的原始问题。
#include <boost/python.hpp>
#include <cassert>
#include <memory> // std::unique_ptr
// Mock legacy API.
struct container
{
container() {}
container(boost::python::object) {}
container(const container&)
{
// For this example, guarantee copy is not made.
assert(false);
}
};
struct my_object
{
int evaluate(container&, double&, container&, double&) { return 42; }
};
/// @brief Transfer ownership to a Python object. If the transfer fails,
/// then object will be destroyed and an exception is thrown.
template <typename T>
boost::python::object transfer_to_python(T* t)
{
// Transfer ownership to a smart pointer, allowing for proper cleanup
// incase Boost.Python throws.
std::unique_ptr<T> ptr(t);
// Use the manage_new_object generator to transfer ownership to Python.
namespace python = boost::python;
typename python::manage_new_object::apply<T*>::type converter;
// Transfer ownership to the Python handler and release ownership
// from C++.
python::handle<> handle(converter(*ptr));
ptr.release();
return python::object(handle);
}
// API wrapper.
boost::python::tuple my_object_compute(
my_object& self, container& x0, double t0)
{
auto x1 = std::make_unique<container>();
double t1 = 21;
int stat = self.evaluate(x0, t0, *x1, t1);
return boost::python::make_tuple(stat, t1, transfer_to_python(x1.release()));
}
BOOST_PYTHON_MODULE(example)
{
namespace python = boost::python;
python::class_<container>("Container")
.def(python::init<python::object>())
;
python::class_<my_object>("MyObject")
.def("compute", &my_object_compute)
;
}
交互使用:
>>> import example
>>> my_object = example.MyObject()
>>> status, t, x = my_object.compute(example.Container([1, 2, 3]), 4)
>>> assert(status == 42)
>>> assert(t == 21)
>>> assert(isinstance(x, example.Container))