你可以痛饮的boost ::可选<&GT ;? [英] Can you SWIG a boost::optional<>?

问题描述

我一直使用痛饮成功地构建一个包装器界面，使在C＃中使用我的C ++库。最近我暴露了一些的boost ::可选<> 对象和痛饮是与他们有问题。是否有处理这个标准的方式吗？一定有人以前碰到这个...

I've been using SWIG successfully to build a wrapper interface to make my C++ libraries available in C#. Recently I exposed some boost::optional<> objects and SWIG is having problems with them. Is there a standard way to deal with this? Someone must have run into this before...

推荐答案

由于痛饮不理解升压型，typemaps必须写。下面是升压一对typemaps的::可选< INT方式>

Since SWIG doesn't understand boost types, typemaps have to be written. Here's a pair of typemaps for boost::optional<int>.

从Python中，无或整数可以传递给函数：

From Python, None or an integer can be passed into a function:

%typemap(in) boost::optional<int> %{
    if($input == Py_None)
        $1 = boost::optional<int>();
    else
        $1 = boost::optional<int>(PyLong_AsLong($input));
%}

一个返回的boost ::可选< INT> 将被转换为无或Python的整数：

A returned boost::optional<int> will be converted to a None or a Python integer:

%typemap(out) boost::optional<int> %{
    if($1)
        $result = PyLong_FromLong(*$1);
    else
    {
        $result = Py_None;
        Py_INCREF(Py_None);
    }
%}

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