我是一个闭源C ++库,它提供的头文件的代码相当于:
class CSomething
{
public:
void getParams( unsigned char & u8OutParamOne,
unsigned char & u8OutParamTwo ) const;
private:
unsigned char u8OutParamOne_,
unsigned char u8OutParamTwo_,
};
我正在尝试向Python公开,我的包装代码是这样的:
BOOST_PYTHON_MODULE(MySomething)
{
class_<CSomething>("CSomething", init<>())
.def("getParams", &CSomething::getParams,(args("one", "two")))
}
现在我正在尝试在Python中使用它,它失败了:
one, two = 0, 0
CSomething.getParams(one, two)
结果是:
ArgumentError: Python argument types in
CSomething.getParams(CSomething, int, int)
did not match C++ signature:
getParams(CSomething {lvalue}, unsigned char {lvalue} one, unsigned char {lvalue} two)
我需要在Boost.Python包装器代码或Python代码中进行哪些更改才能使其工作?如何添加一些Boost.Python魔法来自动将PyInt
转换为unsigned char
,反之亦然?
答案 0 :(得分:1)
Boost.Python
抱怨缺少lvalue
参数,这个概念在Python中不存在:
def f(x):
x = 1
y = 2
f(y)
print(y) # Prints 2
x
函数的f
参数不是类似C ++的参考。在C ++中,输出是不同的:
void f(int &x) {
x = 1;
}
void main() {
int y = 2;
f(y);
cout << y << endl; // Prints 1.
}
你有几个选择:
a)包装CSomething.getParams
函数以返回新参数值的元组:
one, two = 0, 0
one, two = CSomething.getParams(one, two)
print(one, two)
b)包装CSomething.getParams
函数以接受类实例作为参数:
class GPParameter:
def __init__(self, one, two):
self.one = one
self.two = two
p = GPParameter(0, 0)
CSomething.getParams(p)
print(p.one, p.two)