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[英]C++ and Boost.Python - how to expose variable to python and update it in loop?
[英]How to expose a c++ function taking variable arguments in boost python
我有一个c ++函数接受可变数量的参数。
char const* Fun(int num, ...)
{
....//does some processing on the arguments passed
}
用于公开此功能的Boost Python代码编写为:
using namespace boost::python;
BOOST_PYTHON_MODULE( lib_boost )
{
def( "Fun", Fun );
}
在编译此代码时出现以下错误
在/boost_1_42_0/boost/python/data_members.hpp:15、/boost_1_42_0/boost/python/class.hpp:17、/boost_1_42_0/boost/python.hpp:18、Lib_boost.h:3,来自Lib_boost.cpp:1:/boost_1_42_0/boost/python/make_function.hpp:在功能'boost :: python :: api :: object boost :: python :: make_function(F)[with F = const char *( ) (int,...)]':/ boost_1_42_0 / boost / python / def.hpp:82:
从'boost :: python :: api :: object boost :: python :: detail :: make_function1(T,...)[with T = const char ( )(int,...)]实例化'/ boost_1_42_0 / boost / python / def.hpp:91:从'void boost :: python :: def(const char ,Fn)[与Fn = const char *( )(int,...)]实例化“ Lib_boost.cpp:540 :从此处实例化/boost_1_42_0/boost/python/make_function.hpp:104:错误:从'const char ( )(int,...)'到'const char ( )(int)/ boost_1_42_0 / boost / python的 无效转换 /make_function.hpp:104:错误:
初始化'boost :: mpl :: vector2 boost :: python :: detail :: get_signature(RT( )(T0),void *)[with RT = const char *,T0 = int]的参数1
从上面的错误信息中我的理解是boost python无法识别带有可变参数的函数(从'const char *( )(int,...)'到'const char (*)(int)'的无效转换)
公开带有固定/已知参数集的函数与采用可变参数的函数不同。 如何公开带有可变参数的函数?
我发现处理可变参数的最佳方法是使用raw_function
。 通过这种方式,您可以完全控制将C ++参数转换为Python对象:
包装器:
using namespace boost::python;
object fun(tuple args, dict kwargs)
{
char* returned_value;
for(int i = 0; i < len(args); ++i) {
// Extract the args[i] into a C++ variable,
// build up your argument list
}
// build your parameter list from args and kwargs
// and pass it to your variadic c++ function
return str(returned_value);
}
声明:
def("fun", raw_function(fun, 1) );
raw_function
有两个参数:函数指针和最小数量的参数。
如果您使用的是c ++ 11,则可以执行以下操作(在g ++-4.8.2上测试)
#include <boost/python.hpp>
#include <boost/python/list.hpp>
#include <vector>
#include <string>
#include <cstdarg>
#include <cassert>
using namespace boost::python;
template <int... Indices>
struct indices
{
using next = indices<Indices..., sizeof...(Indices)>;
};
template <int N>
struct build_indices
{
using type = typename build_indices<N-1>::type::next;
};
template <>
struct build_indices<0>
{
using type = indices<>;
};
template <int N>
using BuildIndices = typename build_indices<N>::type;
template <int num_args>
class unpack_caller
{
private:
template <typename FuncType, int... I>
char * call(FuncType &f, std::vector<char*> &args, indices<I...>)
{
return f(args.size(), args[I]...);
}
public:
template <typename FuncType>
char * operator () (FuncType &f, std::vector<char*> &args)
{
assert( args.size() <= num_args );
return call(f, args, BuildIndices<num_args>{});
}
};
//This is your function that you wish to call from python
char * my_func( int a, ... )
{
//do something ( this is just a sample )
static std::string ret;
va_list ap;
va_start (ap, a);
for( int i = 0; i < a; ++i)
{
ret += std::string( va_arg (ap, char * ) );
}
va_end (ap);
return (char *)ret.c_str();
}
std::string my_func_overload( list & l )
{
extract<int> str_count( l[0] );
if( str_count.check() )
{
int count = str_count();
std::vector< char * > vec;
for( int index = 1; index <= count; ++index )
{
extract< char * > str( l[index] );
if( str.check() )
{
//extract items from list and build vector
vec.push_back( str() );
}
}
//maximum 20 arguments will be processed.
unpack_caller<20> caller;
return std::string( caller( my_func, vec ) );
}
return std::string("");
}
BOOST_PYTHON_MODULE(my_module)
{
def("my_func", my_func_overload )
;
}
在python中:
Python 2.7.6 (default, Mar 22 2014, 22:59:38)
[GCC 4.8.2] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import my_module as m
>>> m.my_func([5, "my", " first", " five", " string", " arguments"])
'my first five string arguments'
>>>
在此示例中,“ char * my_func(int a,...)”仅连接所有字符串参数并返回结果字符串。
只要知道最大数量是多少,就可以通过将参数视为可选参数来实现。 看到这里: https : //wiki.python.org/moin/boost.python/FunctionOverloading
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