在boost :: python的add_property中使用c ++ 11 lambda作为访问器函数（get_signature失败，使用lambda） [英] Using c++11 lambda as accessor function in boost::python's add_property (get_signature fails with lambda)

问题描述

我试图使用c ++ 11 lambdas作为 boost :: python中的访问器函数' add_property ，下面是一些东西（在这个例子中lambda不是严格需要的，但是在lambda内部发生的更复杂的事情，例如验证将需要lambda）：

I am trying to use c++11 lambdas as accessor functions in boost::python's add_property, something along the following (the lambda is not strictly needed in this example, but will be needed for more complicated things happening inside the lambda, such as validation):

#include<boost/python.hpp>

struct A{
  A(): a(2){};
  int a;
};

BOOST_PYTHON_MODULE(boost_python_lambda)
{
  boost::python::class_<A>("A")
    // .def_readonly("a",&A::a) // the classical way: works fine 
    .add_property("a",[](const A& a){return a.a;})
  ;
}

$ b < > -std = c ++ 11 （结果与g ++ 4.7相同），我得到这个错误：

However, compiling with clang++ (ver. 3.2) and -std=c++11 (the result is the same with g++ 4.7), I get this error:

/usr/include/boost/python/class.hpp:442:66: error: no matching function for call to 'get_signature'
        return python::make_function(f, default_call_policies(), detail::get_signature(f, (T*)0));
                                                                 ^~~~~~~~~~~~~~~~~~~~~
/usr/include/boost/python/class.hpp:422:22: note: in instantiation of function template specialization 'boost::python::class_<A,
      boost::python::detail::not_specified, boost::python::detail::not_specified,
      boost::python::detail::not_specified>::make_fn_impl<A, <lambda at boost_python_lambda.cpp:12:21> >' requested here
        return this->make_fn_impl(
                     ^
/usr/include/boost/python/class.hpp:309:40: note: in instantiation of function template specialization 'boost::python::class_<A,
      boost::python::detail::not_specified, boost::python::detail::not_specified,
      boost::python::detail::not_specified>::make_getter<<lambda at boost_python_lambda.cpp:12:21> >' requested here
        base::add_property(name, this->make_getter(fget), docstr);
                                       ^
boost_python_lambda.cpp:12:4: note: in instantiation of function template specialization 'boost::python::class_<A,
      boost::python::detail::not_specified, boost::python::detail::not_specified,
      boost::python::detail::not_specified>::add_property<<lambda at boost_python_lambda.cpp:12:21> >' requested here
                .add_property("a",[](const A& a){return a.a;})
                 ^

我尝试在 std :: function< int（const A&）>（...）中封装lambda，但是没有帮助参数扣除。任何想法？

I tried wrapping the lambda in std::function<int(const A&)>(...), but that did not help with the argument deduction. Any idea?

推荐答案

在这两年后，Boost.Python确实不支持包装函数对象。但你的lambda没有捕获任何东西。因此，它可以显式转换为函数指针：

Hopping in here two years later, Boost.Python indeed does not support wrapping function objects. But your lambda does not capture anything. As such, it can be explicitly convertible to a function pointer:

BOOST_PYTHON_MODULE(boost_python_lambda)
{
  boost::python::class_<A>("A")
    // .def_readonly("a",&A::a) // the classical way: works fine 
    .add_property("a", +[](const A& a){return a.a;})
                       ↑↑↑
  ;
}

所有你需要的是 + 。

这篇关于在boost :: python的add_property中使用c ++ 11 lambda作为访问器函数（get_signature失败，使用lambda）的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！