我目前正在使用Boost.Python，希望能帮助解决一个棘手的问题。
上下文
当C ++方法/函数暴露给Python时，它需要释放GIL（全局解释器锁）以让其他线程使用解释器。这样，当python代码调用C ++函数时，解释器可以被其他线程使用。 目前，每个C ++函数都是这样的：
// module.cpp
int myfunction(std::string question)
{
ReleaseGIL unlockGIL;
return 42;
}
要将它传递给boost python，我会这样做：
// python_exposure.cpp
BOOST_PYTHON_MODULE(PythonModule)
{
def("myfunction", &myfunction);
}
问题
此方案运行正常，但这意味着module.cpp依赖Boost.Python没有充分理由。理想情况下，只有python_exposure.cpp应取决于Boost.Python。
解决方案吗
我的想法是使用Boost.Function来包装函数调用，如下所示：
// python_exposure.cpp
BOOST_PYTHON_MODULE(PythonModule)
{
def("myfunction", wrap(&myfunction));
}
此处wrap负责在致电myfunction期间解锁GIL。这种方法的问题是wrap需要与myfunction具有相同的签名，这几乎意味着重新实现Boost.Function ......
如果有人对此问题有任何建议，我将非常感激。
答案 0 :(得分：12)
将仿函数公开为方法不是officially supported。支持的方法是公开委托给成员函数的非成员函数。但是，这可能会导致大量的样板代码。
据我所知，Boost.Python的实现没有明确排除仿函数，因为它允许python::object的实例作为方法公开。但是，Boost.Python确实对作为方法公开的对象类型提出了一些要求：
o可以称为o(a1, a2, a3)。boost::python::detail::get_signature()函数来获取此元数据。元数据在内部用于设置正确的调用，以及从Python调度到C ++。后一个要求是复杂的地方。由于某些原因我不能立即清楚，Boost.Python通过qualified-id调用get_signature()，防止参数依赖查找。因此，必须在调用模板的定义上下文之前声明get_signature()的所有候选项。例如，get_signature()所考虑的唯一重载是在调用它的模板定义之前声明的重载，例如class_，def()和make_function()。为了解释这种行为，在Boost.Python中启用仿函数时，必须在包含Boost.Python之前提供get_signature()重载，或者明确地向make_function()提供表示签名的元序列。
让我们通过一些启用仿函数支持的示例，以及提供支持守卫的仿函数。我选择不使用C ++ 11功能。因此，可以使用可变参数模板减少一些样板代码。此外，所有示例都将使用提供两个非成员函数的相同模型和具有两个成员函数的spam类：
/// @brief Mockup class with member functions.
class spam
{
public:
void action()
{
std::cout << "spam::action()" << std::endl;
}
int times_two(int x)
{
std::cout << "spam::times_two()" << std::endl;
return 2 * x;
}
};
// Mockup non-member functions.
void action()
{
std::cout << "action()" << std::endl;
}
int times_two(int x)
{
std::cout << "times_two()" << std::endl;
return 2 * x;
}
boost::function 将preferred syntax用于Boost.Function时，可以使用Boost.FunctionTypes将签名分解为满足Boost.Python要求的元数据。这是一个完整的示例，使boost::function仿函数可以作为Boost.Python方法公开：
#include <iostream>
#include <boost/function.hpp>
#include <boost/function_types/components.hpp>
namespace boost {
namespace python {
namespace detail {
// get_signature overloads must be declared before including
// boost/python.hpp. The declaration must be visible at the
// point of definition of various Boost.Python templates during
// the first phase of two phase lookup. Boost.Python invokes the
// get_signature function via qualified-id, thus ADL is disabled.
/// @brief Get the signature of a boost::function.
template <typename Signature>
inline typename boost::function_types::components<Signature>::type
get_signature(boost::function<Signature>&, void* = 0)
{
return typename boost::function_types::components<Signature>::type();
}
} // namespace detail
} // namespace python
} // namespace boost
#include <boost/python.hpp>
/// @brief Mockup class with member functions.
class spam
{
public:
void action()
{
std::cout << "spam::action()" << std::endl;
}
int times_two(int x)
{
std::cout << "spam::times_two()" << std::endl;
return 2 * x;
}
};
// Mockup non-member functions.
void action()
{
std::cout << "action()" << std::endl;
}
int times_two(int x)
{
std::cout << "times_two()" << std::endl;
return 2 * x;
}
BOOST_PYTHON_MODULE(example)
{
namespace python = boost::python;
// Expose class and member-function.
python::class_<spam>("Spam")
.def("action", &spam::action)
.def("times_two", boost::function<int(spam&, int)>(
&spam::times_two))
;
// Expose non-member function.
python::def("action", &action);
python::def("times_two", boost::function<int()>(
boost::bind(×_two, 21)));
}
及其用法：
>>> import example
>>> spam = example.Spam()
>>> spam.action()
spam::action()
>>> spam.times_two(5)
spam::times_two()
10
>>> example.action()
action()
>>> example.times_two()
times_two()
42
当提供将调用成员函数的仿函数时，提供的签名需要是非成员函数等效函数。在这种情况下，int(spam::*)(int)变为int(spam&, int)。
// ...
.def("times_two", boost::function<int(spam&, int)>(
&spam::times_two))
;
此外，参数可以绑定到boost::bind的仿函数。例如，调用example.times_two()不必提供参数，因为21已绑定到仿函数。
python::def("times_two", boost::function<int()>(
boost::bind(×_two, 21)));
扩展上面的例子，可以启用自定义函子类型与Boost.Python一起使用。让我们创建一个名为guarded_function的仿函数，它将使用RAII，只在RAII对象的生命周期内调用包装函数。
/// @brief Functor that will invoke a function while holding a guard.
/// Upon returning from the function, the guard is released.
template <typename Signature,
typename Guard>
class guarded_function
{
public:
typedef typename boost::function_types::result_type<Signature>::type
result_type;
template <typename Fn>
guarded_function(Fn fn)
: fn_(fn)
{}
result_type operator()()
{
Guard g;
return fn_();
}
// ... overloads for operator()
private:
boost::function<Signature> fn_;
};
guarded_function提供与Python with语句类似的语义。因此，为了与Boost.Python API名称选择保持一致，with() C ++函数将提供一种创建仿函数的方法。
/// @brief Create a callable object with guards.
template <typename Guard,
typename Fn>
boost::python::object
with(Fn fn)
{
return boost::python::make_function(
guarded_function<Guard, Fn>(fn), ...);
}
这允许暴露函数，这些函数将以非侵入方式与警卫一起运行：
class no_gil; // Guard
// ...
.def("times_two", with<no_gil>(&spam::times_two))
;
此外，with()函数提供了推断函数签名的功能，允许将元数据签名显式提供给Boost.Python，而不必重载boost::python::detail::get_signature()。
以下是使用两种RAII类型的完整示例：
no_gil：在构造函数中释放GIL，并在析构函数中重新获取GIL。echo_guard：在构造函数和析构函数中打印。#include <iostream>
#include <boost/function.hpp>
#include <boost/function_types/components.hpp>
#include <boost/function_types/function_type.hpp>
#include <boost/function_types/result_type.hpp>
#include <boost/python.hpp>
#include <boost/tuple/tuple.hpp>
namespace detail {
/// @brief Functor that will invoke a function while holding a guard.
/// Upon returning from the function, the guard is released.
template <typename Signature,
typename Guard>
class guarded_function
{
public:
typedef typename boost::function_types::result_type<Signature>::type
result_type;
template <typename Fn>
guarded_function(Fn fn)
: fn_(fn)
{}
result_type operator()()
{
Guard g;
return fn_();
}
template <typename A1>
result_type operator()(A1 a1)
{
Guard g;
return fn_(a1);
}
template <typename A1, typename A2>
result_type operator()(A1 a1, A2 a2)
{
Guard g;
return fn_(a1, a2);
}
private:
boost::function<Signature> fn_;
};
/// @brief Provides signature type.
template <typename Signature>
struct mpl_signature
{
typedef typename boost::function_types::components<Signature>::type type;
};
// Support boost::function.
template <typename Signature>
struct mpl_signature<boost::function<Signature> >:
public mpl_signature<Signature>
{};
/// @brief Create a callable object with guards.
template <typename Guard,
typename Fn,
typename Policy>
boost::python::object with_aux(Fn fn, const Policy& policy)
{
// Obtain the components of the Fn. This will decompose non-member
// and member functions into an mpl sequence.
// R (*)(A1) => R, A1
// R (C::*)(A1) => R, C*, A1
typedef typename mpl_signature<Fn>::type mpl_signature_type;
// Synthesize the components into a function type. This process
// causes member functions to require the instance argument.
// This is necessary because member functions will be explicitly
// provided the 'self' argument.
// R, A1 => R (*)(A1)
// R, C*, A1 => R (*)(C*, A1)
typedef typename boost::function_types::function_type<
mpl_signature_type>::type signature_type;
// Create a callable boost::python::object that delegates to the
// guarded_function.
return boost::python::make_function(
guarded_function<signature_type, Guard>(fn),
policy, mpl_signature_type());
}
} // namespace detail
/// @brief Create a callable object with guards.
template <typename Guard,
typename Fn,
typename Policy>
boost::python::object with(const Fn& fn, const Policy& policy)
{
return detail::with_aux<Guard>(fn, policy);
}
/// @brief Create a callable object with guards.
template <typename Guard,
typename Fn>
boost::python::object with(const Fn& fn)
{
return with<Guard>(fn, boost::python::default_call_policies());
}
/// @brief Mockup class with member functions.
class spam
{
public:
void action()
{
std::cout << "spam::action()" << std::endl;
}
int times_two(int x)
{
std::cout << "spam::times_two()" << std::endl;
return 2 * x;
}
};
// Mockup non-member functions.
void action()
{
std::cout << "action()" << std::endl;
}
int times_two(int x)
{
std::cout << "times_two()" << std::endl;
return 2 * x;
}
/// @brief Guard that will unlock the GIL upon construction, and
/// reacquire it upon destruction.
struct no_gil
{
public:
no_gil() { state_ = PyEval_SaveThread();
std::cout << "no_gil()" << std::endl; }
~no_gil() { std::cout << "~no_gil()" << std::endl;
PyEval_RestoreThread(state_); }
private:
PyThreadState* state_;
};
/// @brief Guard that prints to std::cout.
struct echo_guard
{
echo_guard() { std::cout << "echo_guard()" << std::endl; }
~echo_guard() { std::cout << "~echo_guard()" << std::endl; }
};
BOOST_PYTHON_MODULE(example)
{
namespace python = boost::python;
// Expose class and member-function.
python::class_<spam>("Spam")
.def("action", &spam::action)
.def("times_two", with<no_gil>(&spam::times_two))
;
// Expose non-member function.
python::def("action", &action);
python::def("times_two", with<boost::tuple<no_gil, echo_guard> >(
×_two));
}
及其用法：
>>> import example
>>> spam = example.Spam()
>>> spam.action()
spam::action()
>>> spam.times_two(5)
no_gil()
spam::times_two()
~no_gil()
10
>>> example.action()
action()
>>> example.times_two(21)
no_gil()
echo_guard()
times_two()
~echo_guard()
~no_gil()
42
请注意如何使用容器类型（例如boost::tuple：
python::def("times_two", with<boost::tuple<no_gil, echo_guard> >(
×_two));
在Python中调用时，example.times_two(21)会产生以下输出：
no_gil()
echo_guard()
times_two()
~echo_guard()
~no_gil()
42
答案 1 :(得分：2)
如果有人感兴趣，我在使用他最后的工作示例时遇到了 Tanner Sansbury 代码的小问题。出于某种原因，我仍然遇到他提到的关于在最终生成的boost::function中签名错误的问题：
// example for spam::times_two:
// correct signature (manual)
int (spam::*, int)
// wrong signature (generated in the `guarded_function` wrapper)
int (spam&, int)
即使重载boost::python::detail::get_signature()。对此负责是boost::function_types::components;它有一个默认的模板参数ClassTranform = add_reference<_>，用于创建此类引用。要解决这个问题，我只需更改mpl_signature结构如下：
// other includes
# include <boost/type_traits/add_pointer.hpp>
# include <boost/mpl/placeholders.hpp>
template <typename Signature> struct mpl_signature
{
typedef typename boost::function_types::components<Signature,
boost::add_pointer<boost::mpl::placeholders::_> >::type type;
};
template <typename Signature> struct mpl_signature<boost::function<Signature> >
{
typedef typename boost::function_types::components<Signature>::type type;
};
现在一切都像魅力一样。
如果有人可以确认这确实是正确的解决方案，我会感兴趣：）