mongocxx 3.x insert will throw exception when duplicate key found

[li ning]

Not way to disable exception? Why not set error code in return value? Even the return type is a dedicated type "result"?

[Andrew Morrow]

The C++11 driver throws exceptions to indicate errors that the user must handle. Exceptions require the programmer to deal with them - return values may be ignored.

Note that in C++11 (or particularly C++14) you can easily transmute a throwing function into one that returns values, if you are so inclined:

const auto make_nothrow = [](auto callable) {

    return [callable(std::move(callable))](auto ...args) noexcept {

        try {

            return std::make_tuple(

                std::experimental::make_optional(

                    callable(args...)

                ),

                std::exception_ptr()

            );

        }

        catch(...) {

            return std::make_tuple(

                std::experimental::optional<decltype(callable(args...))>(),

                std::current_exception()

            );

        }

    };

};

double throws(int x, double y) {

    if (x != y)

        throw std::runtime_error("I'm an exception");

    return x + y;

}

int main(int argc, char* argv[]) {

    auto doesnt_throw = make_nothrow(throws);

    auto result1 = doesnt_throw(argc, 1.0);

    std::cout << *std::get<0>(result1) << std::endl;

    auto result2 = doesnt_throw(argc, 2.0);

    return EXIT_SUCCESS;

}

I've used C++14 to reduce the template verbiage here, and std::optional and std::tuple here to simulate a variant type, but if you have a proper variant like boost::variant, you can do even better.

Thanks,

Andrew

On Wed, Apr 20, 2016 at 2:59 AM, li ning <li.ni...@gmail.com> wrote:

Not way to disable exception? Why not set error code in return value? Even the return type is a dedicated type "result"?

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