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by Night_Thastus 934 days ago
Strange.

How is that different from say, a function call? A function may look like a single O(1) operation from the input/output/name, but actually do something much more complex. That seems like the same thing to me, and very common. (and frankly I'm not sure that could even be avoided)
1 comments
klyrs 934 days ago
I didn't mean to volunteer to defend this choice, but without investigating a function you can't really support an opinion about its runtime. A language can make such promises about its basic syntax however.

Perhaps I'll rephrase how I understand the philosophy: if it's a function call, it should look like a function call. Operator overloading breaks that.

That said, this isn't my hill to die on.

Edit to clarify my final sentence there: I have zero interest in debating this any further. Pixelpoet, if you're going to be so fussy, go read Harvey and van der Hoeven and stop trying to win language fights, they're tedious.

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pixelpoet 934 days ago
I appreciate your being such a good devil's advocate, however as ForkMeOnTinder points out, there is already the super complicated[0] overloaded O(N^1.58+) operator for arbitrary precision integer multiplication, easily argued to be vanishingly less useful than simple vector operators, particularly if they were well-expressed at the IR level and well-mapped to modern hardware / instructions.

The ask isn't for general operator overloading (I'm also in favour of not having that), rather just not stopping native algebraic type support at scalars; the C function atan2(y, x) basically just wants to give you the complex argument, for example. Really it would just do so much to unify and simplify everything, besides being able to write vector stuff in a sane way; if every rando has to write their own vector and complex number classes, I'm much less likely to vouch for its correctness.

[0] I've recently been looking into Karatsuba multiplication to reduce it from O(N^2) to O(N^1.58): https://en.wikipedia.org/wiki/Karatsuba_algorithm

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Night_Thastus 934 days ago
To me it does seem a weird hill to die on. If I'm using an operator on a non-primitive/non-trivial type - I'm going to consider the possibility that it does something more complex. It would be strange not to.
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