如何在boost.python中指定命名参数的值? [英] How can I specify the value of a named argument in boost.python?

问题描述

我想将python编写的函数嵌入到c ++代码中.
我的python代码是:test.py

i want to embed a function written in python into c++ code.
My python code is:test.py

def func(x=None, y=None, z=None):  
  print x,y,z  

我的c ++代码是:

module = import("test");  
namespace = module.attr("__dict__");  

//then i want to know how to pass value 'y' only.  
module.attr("func")("y=1") // is that right?

推荐答案

我不确定Boost.Python是否实现了声明的**解除引用运算符，但是您仍然可以使用Python C-API执行您所使用的方法如此处所述进行.

I'm not sure Boost.Python implements the ** dereference operator as claimed, but you can still use the Python C-API to execute the method you are intested on, as described here.

这是解决方案的原型:

//I'm starting from where you should change
boost::python::object callable = module.attr("func");

//Build your keyword argument dictionary using boost.python
boost::python::dict kw;
kw["x"] = 1;
kw["y"] = 3.14;
kw["z"] = "hello, world!";

//Note: This will return a **new** reference
PyObject* c_retval = PyObject_Call(callable.ptr(), NULL, kw.ptr());

//Converts a new (C) reference to a formal boost::python::object
boost::python::object retval(boost::python::handle<>(c_retval));

将返回值从PyObject_Call转换为正式的boost::python::object之后，您可以从函数中返回它，也可以忘了它，并且PyObject_Call返回的新引用将被自动删除.

After you have converted the return value from PyObject_Call to a formal boost::python::object, you can either return it from your function or you can just forget it and the new reference returned by PyObject_Call will be auto-deleted.

有关将PyObject*包装为boost::python::object的更多信息，请参阅Boost.Python教程. 更精确地说，链接，页面的结尾.

For more information about wrapping PyObject* as boost::python::object, have a look at the Boost.Python tutorial. More precisely, at this link, end of the page.

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