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Home/ Questions/Q 8320771
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Editorial Team
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Editorial Team
Asked: June 8, 20262026-06-08T22:46:29+00:00 2026-06-08T22:46:29+00:00

The accepted answer to boost::python Export Custom Exception shows how to export a custom

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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.

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  1. Editorial Team
    Editorial Team
    2026-06-08T22:46:31+00:00Added an answer on June 8, 2026 at 10:46 pm

    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());
    }
    

    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
    

    but the exception will also be caught by

    try:
        ...
    except Exception:
        ...
    
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