Question

我目前正在使用Boost.Python为Python编写C ++扩展。此扩展中的函数可能会生成一个包含错误信息的异常（除了描述发生的事件的人类可读字符串之外）。我希望我可以将这个异常导出到Python，这样我就能抓住它并用额外的信息做些事情。

例如：

import my_cpp_module
try:
    my_cpp_module.my_cpp_function()
except my_cpp_module.MyCPPException, e:
    print e.my_extra_data

不幸的是，Boost.Python似乎将所有C ++异常（即std::exception的子类）转换为RuntimeError。我意识到Boost.Python允许实现自定义异常转换，但是需要使用PyErr_SetObject，其中PyObject*（对于异常的类型）和PyObject*（对于异常的值）） - 我不知道如何从我的Boost.Python类中获取。也许有一种方法（这将是伟大的），我还没有找到。否则有人知道如何导出自定义C ++异常，以便我可以在Python中捕获它吗？

Answer 1

解决方案是像任何普通的C ++类一样创建异常类

class MyCPPException : public std::exception {...}

诀窍是所有boost :: python :: class_实例都包含对对象类型的引用，可以通过它们的ptr（）函数访问它。您可以使用boost :: python注册类，因为：

class_<MyCPPException> myCPPExceptionClass("MyCPPException"...);
PyObject *myCPPExceptionType=myCPPExceptionClass.ptr();
register_exception_translator<MyCPPException>(&translateFunc);

最后，当您将C ++异常翻译为Python异常时，请执行以下操作：

void translate(MyCPPException const &e)
{
    PyErr_SetObject(myCPPExceptionType, boost::python::object(e).ptr());
}

以下是一个完整的工作示例：

#include <boost/python.hpp>
#include <assert.h>
#include <iostream>

class MyCPPException : public std::exception
{
private:
  std::string message;
  std::string extraData;
public:
  MyCPPException(std::string message, std::string extraData)
  {
    this->message = message;
    this->extraData = extraData;
  }
  const char *what() const throw()
  {
    return this->message.c_str();
  }
  ~MyCPPException() throw()
  {
  }
  std::string getMessage()
  {
    return this->message;
  }
  std::string getExtraData()
  {
    return this->extraData;
  }
};

void my_cpp_function(bool throwException)
{
  std::cout << "Called a C++ function." << std::endl;
  if (throwException)
    {
      throw MyCPPException("Throwing an exception as requested.",
               "This is the extra data.");
    }
}

PyObject *myCPPExceptionType = NULL;

void translateMyCPPException(MyCPPException const &e)
{
  assert(myCPPExceptionType != NULL);
  boost::python::object pythonExceptionInstance(e);
  PyErr_SetObject(myCPPExceptionType, pythonExceptionInstance.ptr());
}

BOOST_PYTHON_MODULE(my_cpp_extension)
{
  boost::python::class_<MyCPPException>
    myCPPExceptionClass("MyCPPException",
            boost::python::init<std::string, std::string>());
  myCPPExceptionClass.add_property("message", &MyCPPException::getMessage)
    .add_property("extra_data", &MyCPPException::getExtraData);
  myCPPExceptionType = myCPPExceptionClass.ptr();
  boost::python::register_exception_translator<MyCPPException>
    (&translateMyCPPException);
  boost::python::def("my_cpp_function", &my_cpp_function);
}

以下是调用扩展名的Python代码：

import my_cpp_extension
try:
    my_cpp_extension.my_cpp_function(False)
    print 'This line should be reached as no exception should be thrown.'
except my_cpp_extension.MyCPPException, e:
    print 'Message:', e.message
    print 'Extra data:',e.extra_data

try:
    my_cpp_extension.my_cpp_function(True)
    print ('This line should not be reached as an exception should have been' +
       'thrown by now.')
except my_cpp_extension.MyCPPException, e:
    print 'Message:', e.message
    print 'Extra data:',e.extra_data

Answer 2

Jack Edmonds给出的答案定义了一个Python“异常”类，它不继承Exception（或任何其他内置的Python异常类）。所以虽然它可以被抓住

except my_cpp_extension.MyCPPException as e:
    ...

它不能被通常的捕捉所有

抓住

except Exception as e:
    ...

Here是如何创建继承Exception的自定义Python异常类。

Answer 3

感谢可变参数模板和广义lambda捕获，我们可以将Jack Edmond's answer折叠成更易于管理的东西，并隐藏用户的所有信息：

template <class E, class... Policies, class... Args>
py::class_<E, Policies...> exception_(Args&&... args) {
    py::class_<E, Policies...> cls(std::forward<Args>(args)...);
    py::register_exception_translator<E>([ptr=cls.ptr()](E const& e){
        PyErr_SetObject(ptr, py::object(e).ptr());
    });
    return cls;
}

要将MyCPPException公开为例外，您只需将绑定中的py::class_更改为exception_：

exception_<MyCPPException>("MyCPPException", py::init<std::string, std::string>())
    .add_property("message", &MyCPPException::getMessage)
    .add_property("extra_data", &MyCPPException::getExtraData)
;

现在我们又回到了Boost.Python的细节：不需要为class_实例命名，不需要额外的PyObject*，并且在某个地方不需要额外的功能。

Answer 4

这是杰克·埃德蒙兹（Jack Edmonds）的解决方案，它使用here的建议移植到Python 3，而该建议本身使用的是here的代码。将它们组装在一起（并稍微现代化C ++代码）可以得到：

limit by

和要测试的python文件：

#include <boost/python.hpp>
#include <assert.h>
#include <iostream>

class MyCPPException : public std::exception
{
public:
    MyCPPException(const std::string &message, const std::string &extraData)
        : message(message), extraData(extraData)
    {
    }
    const char *what() const noexcept override
    {
        return message.c_str();
    }
    std::string getMessage() const
    {
        return message;
    }
    std::string getExtraData() const
    {
        return extraData;
    }
private:
    std::string message;
    std::string extraData;
};

void my_cpp_function(bool throwException)
{
    std::cout << "Called a C++ function." << std::endl;
    if (throwException) {
        throw MyCPPException("Throwing an exception as requested.",
                             "This is the extra data.");
    }
}

static PyObject* createExceptionClass(const char* name, PyObject* baseTypeObj = PyExc_Exception)
{
    using std::string;
    namespace bp = boost::python;

    const string scopeName = bp::extract<string>(bp::scope().attr("__name__"));
    const string qualifiedName0 = scopeName + "." + name;
    PyObject* typeObj = PyErr_NewException(qualifiedName0.c_str(), baseTypeObj, 0);
    if (!typeObj) bp::throw_error_already_set();
    bp::scope().attr(name) = bp::handle<>(bp::borrowed(typeObj));
    return typeObj;
}

static PyObject *pythonExceptionType = NULL;

static void translateMyCPPException(MyCPPException const &e)
{
    using namespace boost;
    python::object exc_t(python::handle<>(python::borrowed(pythonExceptionType)));
    exc_t.attr("cause") = python::object(e); // add the wrapped exception to the Python exception
    exc_t.attr("what") = python::object(e.what()); // for convenience
    PyErr_SetString(pythonExceptionType, e.what()); // the string is used by print(exception) in python
}

BOOST_PYTHON_MODULE(my_cpp_extension)
{
    using namespace boost;
    python::class_<MyCPPException>
            myCPPExceptionClass("MyCPPException",
                                python::init<std::string, std::string>());
    myCPPExceptionClass.add_property("message", &MyCPPException::getMessage)
            .add_property("extra_data", &MyCPPException::getExtraData);

    pythonExceptionType = createExceptionClass("MyPythonException");
    python::register_exception_translator<MyCPPException>(&translateMyCPPException);
    python::def("my_cpp_function", &my_cpp_function);
}

并将其作为标准python异常捕获也可以：

#!/usr/bin/env python3

import my_cpp_extension
try:
    my_cpp_extension.my_cpp_function(False)
    print('This line should be reached as no exception should be thrown.')
except my_cpp_extension.MyPythonException as e:
    print('Message:', e.what)
    print('Extra data:',e.cause.extra_data)

try:
    my_cpp_extension.my_cpp_function(True)
    print ('This line should not be reached as an exception should have been' +
       'thrown by now.')
except my_cpp_extension.MyPythonException as e:
    print('Message:', e.what)
    print('Extra data:',e.cause.extra_data)

boost :: python导出自定义异常

4 个答案: