暴露类的 boost::tuple 部分来提升 python
我一直在试图弄清楚如何在我的类中公开一个 boost::tuple 属性。元组定义如下：
typedef boost::shared_ptr<Action> action_ptr;
typedef boost::tuple<BattleCharacter*, action_ptr > ActionTargetTuple;
它包含一个定义如下的类：
class Action : public Cloneable<Action>
{
public:
//Irrelevant Code Omitted
std::vector<ActionTargetTuple> Targets;
}
当我搜索如何将 boost::tuple 转换为 python 元组时，我看过很多文章，但这不是我想要做的。我希望能够访问 Action 类中存在的元组。 （我知道如何做矢量部分）。
class_<Action, std::auto_ptr<ActionWrapper> >("Action")
.def("Targets", &Action::Targets)
;
我就按照上面简单的暴露一下。我想我也许可以通过下面的一些变体来公开它：
class_<ActionTargetTuple>("ActionTargetTuple")
.def("get", &ActionTargetTuple::get<int>, return_value_policy<reference_existing_object>())
;
然后使用从 python 获取，但如果以这种方式可行，我不确定需要进行什么设置。有谁知道如何做到这一点/可以建议替代方案吗？
谢谢
I've been trying to figure out how to expose a property in my class that is a boost::tuple. The tuple is defined as follows:
typedef boost::shared_ptr<Action> action_ptr;
typedef boost::tuple<BattleCharacter*, action_ptr > ActionTargetTuple;
It's contained with a class defined as follows:
class Action : public Cloneable<Action>
{
public:
//Irrelevant Code Omitted
std::vector<ActionTargetTuple> Targets;
}
I've seen numerous articles while I was searching about how to convert a boost::tuple into a python tuple, but that's not what I'm looking to do. I want to be able to access the tuple as it exists on the Action class. (I know how to do the vector part).
class_<Action, std::auto_ptr<ActionWrapper> >("Action")
.def("Targets", &Action::Targets)
;
I expose it simply as above. I figured I might be able to expose it by some variation on the below:
class_<ActionTargetTuple>("ActionTargetTuple")
.def("get", &ActionTargetTuple::get<int>, return_value_policy<reference_existing_object>())
;
then use get from python, but if it is doable in this way, I'm not sure what the set up needs to be. Does anyone know how to do this/could suggest an alternative?
Thanks
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您可以使用：
在 C++ 中使用 getter/setter 方法创建读写属性
如果您想控制所有权：
You can use:
to make a read-write property using getter/setter methods in c++
If you want to control ownership:
除了使用上一个答案中解释的 add_property 和编写访问器函数之外，您还可以考虑为元组编写转换器（在 boost::tuple 和 boost:: 之间） python::tuple）并直接使用
def_readonly或def_readwrite公开这些属性。如果您有许多这样的属性需要公开，那么这是值得的。This 有一个可以用于 c++→python 转换的模板（使用 boost::tuple 而不是 std::pair），不过除非你使用 c++0x，否则你必须写出不同数量参数的模板。
如果您的属性是读写的，请另外定义 from-python 转换器，您可以在网上找到示例。 这里是我的代码用于定义通用的sequence-std::vector 转换器。在您的情况下，您必须检查 python 对象是否是一个序列，它是否具有正确数量的项目，您可以从每个项目中提取所需的类型；然后返回新的 boost::tuple 对象。
HTH，edx。
PS我发现ackward 已准备好转换器，也许您可以重复使用它。 此处的文档
Besides using
add_propertyas explained in the previous answer, and writing accessor functions, you can consider writing converters for your tuple (betweenboost::tupleandboost::python::tuple) and exposing those attributes directly withdef_readonlyordef_readwrite. It is worth it if you have many such attributes to expose.This has a template you adapt can for c++→python conversion (use boost::tuple instead of std::pair), though unless you go c++0x, you have to write out templates for different number of arguments.
If your property is read-write, additionaly define from-python converter, you find examples on the web. Here is my code I use to define generic sequence-std::vector converter. In your case, you have to check that the python object is a sequence, that it has the right number of items, that you can extract required types from each of them; and then return new boost::tuple object.
HTH, edx.
P.S. I found ackward has the converters ready, perhaps you could just reuse it. Doc here