3 ms·

That's not Turing complete. That's still primitive recursive. You're going out of your way make it loop for a really, really long time, but it's still guarantee

That's not Turing complete. That's still primitive recursive. You're going out of your way make it loop for a really, really long time, but it's still guaranteed to terminate, and thus isn't Turing complete, even if it'd take a really long time to terminate. This isn't some theoretical thing: it's how you can introduce guarantees into the system that it can't get out of control by accident and it also makes it easier to reason about the system. A primitive recursive language gets you most of the benefits of a general recursive one _without_ needing to be _actually_ Turing complete. So this isn't "bad news" for me. I'm saying that I think you should avoid using a general recursive (i.e. Turing complete) language for a build system and instead have guarantees that the system will terminate by using a primitive recursive language of some kind because you literally, as you did, need to go out of your way to get code written in such a fashion to run for an inordinate amount of time. And yet, an inordinate amount of time isn't forever and unlike a program written in an actually Turing complete language, you can use static analysis to detect that kind of thing if only primitive recursion is supported.

gavinhoward 3y ago
> And yet, an inordinate amount of time isn't forever If it's only guaranteed to terminate after the heat death of the universe, it won't terminate, so I would call that "forever." > unlike a program written in an actually Turing complete language, you can use static analysis to detect that kind of thing if only primitive recursion is supported. And this is exactly why I said that break was necessary. Sure your static analysis may detect that the loop goes 2^256 times, which would be "forever," and so it returns the answer, "Does not halt." Ah, but with the break keyword, it still could. In fact, it might always halt rather quickly. Rule of thumb is that if static analysis can be wrong, it is Turing-complete.

gavinhoward 3y ago

> And yet, an inordinate amount of time isn't forever If it's only guaranteed to terminate after the heat death of the universe, it won't terminate, so I would call that "forever." > unlike a program written in an actually Turing complete language, you can use static analysis to detect that kind of thing if only primitive recursion is supported. And this is exactly why I said that break was necessary. Sure your static analysis may detect that the loop goes 2^256 times, which would be "forever," and so it returns the answer, "Does not halt." Ah, but with the break keyword, it still could. In fact, it might always halt rather quickly. Rule of thumb is that if static analysis can be wrong, it is Turing-complete.