我开始学习使用boost python并有一个菜鸟问题。
我想编写一个函数，可以将其参数的生命周期与其结果联系起来，这样当我调用r = func(a)时，如果我仍然有参考，则参数a永远不会被销毁r。该文档建议对此类请求使用return_internal_reference调用策略。但这是否需要r作为a的内部参考，顾名思义？
在下面的（过度简化）示例中，假设我想将输入数组a的生命周期与生成的lambda函数联系起来，该函数不是输入a的内部引用。
#include <functional>
#include <boost/python.hpp>
#include <boost/python/return_internal_reference.hpp>
using namespace std;
using namespace boost::python;
function<float(int)> func(const float* a) {
return [=](int n) { return a[n]; };
}
BOOST_PYTHON_MODULE(test) {
def("func", func, return_internal_reference<1>());
}
我希望能在python中执行以下操作：
f = func(a) # 'a' can be a temporary variable, say returned by another function
f(5) # but 'a' should not be destroyed at this step,
# because its lifetime is tied to 'f'
当我尝试编译上面的代码时，我遇到了下面列出的错误，但是如果我删除了return_internal_reference<1>()调用策略，则代码会成功编译。
我很确定我使用此调用策略是错误的，但我不确定如何使其正确。任何指针都将受到高度赞赏。非常感谢！
$ g++ -std=c++11 -shared Test.cc -I/opt/local/Library/Frameworks/Python.framework/Versions/2.7/include/python2.7 -L/opt/local/lib -lboost_python-mt -lpython2.7 -o test.so
In file included from /opt/local/include/boost/preprocessor/iteration/detail/iter/forward1.hpp:52:0,
from /opt/local/include/boost/python/detail/invoke.hpp:63,
from /opt/local/include/boost/python/detail/caller.hpp:16,
from /opt/local/include/boost/python/object/function_handle.hpp:8,
from /opt/local/include/boost/python/converter/arg_to_python.hpp:19,
from /opt/local/include/boost/python/call.hpp:15,
from /opt/local/include/boost/python/object_core.hpp:14,
from /opt/local/include/boost/python/args.hpp:25,
from /opt/local/include/boost/python.hpp:11,
from Test.cc:2:
/opt/local/include/boost/python/detail/invoke.hpp: In instantiation of 'PyObject* boost::python::detail::invoke(boost::python::detail::invoke_tag_<false, false>, const RC&, F&, AC0&) [with RC = boost::python::detail::reference_existing_object_requires_a_pointer_or_reference_return_type<std::function<float(int)> >; F = std::function<float(int)> (*)(const float*); AC0 = boost::python::arg_from_python<const float*>; PyObject = _object]':
/opt/local/include/boost/python/detail/caller.hpp:223:13: required from 'PyObject* boost::python::detail::caller_arity<1u>::impl<F, Policies, Sig>::operator()(PyObject*, PyObject*) [with F = std::function<float(int)> (*)(const float*); Policies = boost::python::return_internal_reference<>; Sig = boost::mpl::vector2<std::function<float(int)>, const float*>; PyObject = _object]'
/opt/local/include/boost/python/object/py_function.hpp:38:33: required from 'PyObject* boost::python::objects::caller_py_function_impl<Caller>::operator()(PyObject*, PyObject*) [with Caller = boost::python::detail::caller<std::function<float(int)> (*)(const float*), boost::python::return_internal_reference<>, boost::mpl::vector2<std::function<float(int)>, const float*> >; PyObject = _object]'
Test.cc:14:1: required from here
/opt/local/include/boost/python/detail/invoke.hpp:75:82: error: no match for call to '(const boost::python::detail::reference_existing_object_requires_a_pointer_or_reference_return_type<std::function<float(int)> >) (std::function<float(int)>)'
return rc(f( BOOST_PP_ENUM_BINARY_PARAMS_Z(1, N, ac, () BOOST_PP_INTERCEPT) ));
^
In file included from /opt/local/include/boost/python/object/function_handle.hpp:8:0,
from /opt/local/include/boost/python/converter/arg_to_python.hpp:19,
from /opt/local/include/boost/python/call.hpp:15,
from /opt/local/include/boost/python/object_core.hpp:14,
from /opt/local/include/boost/python/args.hpp:25,
from /opt/local/include/boost/python.hpp:11,
from Test.cc:2:
/opt/local/include/boost/python/detail/caller.hpp: In instantiation of 'static const PyTypeObject* boost::python::detail::converter_target_type<ResultConverter>::get_pytype() [with ResultConverter = boost::python::detail::reference_existing_object_requires_a_pointer_or_reference_return_type<std::function<float(int)> >; PyTypeObject = _typeobject]':
/opt/local/include/boost/python/detail/caller.hpp:240:19: required from 'static boost::python::detail::py_func_sig_info boost::python::detail::caller_arity<1u>::impl<F, Policies, Sig>::signature() [with F = std::function<float(int)> (*)(const float*); Policies = boost::python::return_internal_reference<>; Sig = boost::mpl::vector2<std::function<float(int)>, const float*>]'
/opt/local/include/boost/python/object/py_function.hpp:48:35: required from 'boost::python::detail::py_func_sig_info boost::python::objects::caller_py_function_impl<Caller>::signature() const [with Caller = boost::python::detail::caller<std::function<float(int)> (*)(const float*), boost::python::return_internal_reference<>, boost::mpl::vector2<std::function<float(int)>, const float*> >]'
Test.cc:14:1: required from here
/opt/local/include/boost/python/detail/caller.hpp:102:109: error: 'struct boost::python::detail::reference_existing_object_requires_a_pointer_or_reference_return_type<std::function<float(int)> >' has no member named 'get_pytype'
return create_result_converter((PyObject*)0, (ResultConverter *)0, (ResultConverter *)0).get_pytype();
^
答案 0 :(得分：1)
考虑使用with_custodian_and_ward_postcall CallPolicy。此策略允许按值返回返回类型，同时仍将另一个对象的生命周期延长至少与返回对象的生命周期一样长。
BOOST_PYTHON_MODULE(test) {
def("func", func, with_custodian_and_ward_postcall<0, 1>());
}
如return_internal_reference文档中所述，返回的对象引用了现有的内部对象：
return_internal_reference[...]允许在没有复制指示物的情况下安全地返回内部保存的对象的指针和引用。
文档还简要提到了with_custodian_and_ward_postcall的使用。总之，return_internal_reference对所公开的功能有两个值得注意的效果：
owner_arg表示，至少与保管人一样长。由于返回的Python对象引用的是既没有显式所有权也没有共享所有权的现有对象，Boost.Python会执行类型检查以防止创建悬空引用。在示例代码中，func()按值返回一个仿函数，导致编译器错误，暗示返回类型必须是指针或引用：
struct boost::python::detail:: reference_existing_object_requires_a_pointer_or_reference_return_type
要显式控制返回对象的生命周期，应考虑使用return_value_policy和ResultConverterGenerators模型。例如，如果func()通过new()创建的指针返回了一个仿函数，并且希望将对象的所有权传递给Python，同时仍保持托管和区域关系，则可以使用{链接策略{3}}：
BOOST_PYTHON_MODULE(test) {
def("func", func,
return_value_policy<manage_new_object,
with_custodian_and_ward_postcall<0, 1> >());
}
以下是基于原始代码的完整最小示例，其详细输出为policy composition with_custodian_and_ward_postcall行为：
#include <boost/python.hpp>
#include <iostream>
/// @brief Mockup class with verbose construction and destruction.
class foo
{
public:
foo() { std::cout << "foo() " << this << std::endl; }
foo(const foo&) { std::cout << "foo(const foo&) " << this << std::endl; }
~foo() { std::cout << "~foo() " << this << std::endl; }
};
/// @brief Mockup class with verbose construction and destruction.
class bar
{
public:
bar() { std::cout << "bar() " << this << std::endl; }
bar(const bar&) { std::cout << "bar(const bar&) " << this << std::endl; }
~bar() { std::cout << "~bar() " << this << std::endl; }
};
/// @brief Mockup factory function.
foo make_foo(bar& /* unused */)
{
return foo();
}
BOOST_PYTHON_MODULE(example)
{
namespace python = boost::python;
// Do not allow Foo to be explicitly created from its type.
python::class_<foo>("Foo", python::no_init);
python::class_<bar>("Bar", python::init<>());
// Expose make_foo, that returns a foo object when provided a
// bar object. The bar object's lifetime will be extended to
// be at least as long as that of the returned foo object.
python::def("make_foo", &make_foo,
python::with_custodian_and_ward_postcall<
0, // custodian = returned Foo object
1 // ward = provided Bar object
>());
}
交互式使用：
>>> import example
>>> bar = example.Bar()
bar() 0x125ac30
>>> foo = example.make_foo(bar)
foo() 0x7fffa9b5efff
foo(const foo&) 0x7f1fcbe40090
~foo() 0x7fffa9b5efff
>>> bar = None
>>> foo = None
~foo() 0x7f1fcbe40090
~bar() 0x125ac30
请注意，尽管bar变量设置为None，但实际病房对象仍然存活，直到foo监护人被销毁。