boost :: python导出自定义异常并继承自Python的Exception [英] boost::python export custom exception and inherit from Python's Exception

问题描述

boost :: python导出自定义异常的可接受答案显示了如何导出自定义C ++中的异常类，以及 Boost.Python自定义异常类显示了如何导出异常从Python的Exception继承的类.我该怎么办?这暴露了一个异常类，该异常类具有用于检索信息的自定义方法，并且该类也从Python的Exception派生.

The accepted answer to boost::python Export Custom Exception shows how to export a custom exception class from C++, and Boost.Python custom exception class shows how to export an exception class that inherits from Python's Exception. How can I do both? That is expose an exception class that has custom methods to retrieve information and also have that class be derived from Python's Exception.

推荐答案

Jim Bosch在 C ++信号列表，是使用合成，而不是从包装的C ++异常继承.代码必须像此处一样创建Python异常，然后添加包装的C ++作为Python异常的实例变量.

A workable solution, suggested by Jim Bosch on the C++-sig list, is to use composition instead of inheriting from the wrapped C++ exception. The code must create a Python exception as is done here, and then add the wrapped C++ exception as an instance variable of the Python exception.

void translator(const MyCPPException &x) {
    bp::object exc(x); // wrap the C++ exception

    bp::object exc_t(bp::handle<>(bp::borrowed(exceptionType)));
    exc_t.attr("cause") = exc; // add the wrapped exception to the Python exception

    PyErr_SetString(exceptionType, x.what());
}

然后可以像下面这样从Python访问包装的C ++异常:

The wrapped C++ exception can then be accessed from Python like this:

try:
    ...
except MyModule.MyCPPExceptionType as e:
    cause = e.cause # wrapped exception can be accessed here

但该异常也会被

try:
    ...
except Exception:
    ...

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