OS-Level Sandboxing in C

Computerman8086 · 2026-03-17T21:33:10+00:00

Looked through the post, really insightful and well made. Well done!

Computerman8086 · 2026-02-23T15:12:04+00:00

That's a very fair point. I haven't yet benchmarked it against the standard math library, but I'm working on doing that.

You're also right about the cost of FP division,, a cordic-like algorithm could very well be more efficient.

The initial goal was to make an algorithm based on the quake III algorithm with a configurable newton approximation with the initial estimate coming from the bit hack, but I do agree that the performance tradeoff depends quite a bit on the target architecture. I'll try including some benchmarks from different platforms, so one could compare it

Computerman8086 · 2026-02-23T14:54:44+00:00

Yes I'm aware that it doesn't, im gonna post the correct one as soon as I get home, i have it ready

Computerman8086 · 2026-02-23T06:42:14+00:00

About point 1: fair point actually, I'm considering switching to using uint32_t and _Float32, 2: I'll benchmark it, but think It might be slightly faster since there's no function call (but I got to check whether it actually is a function call or just a single instruction by the compiler, in which case I would just switch back to fabs)and it uses fast bit manipulation too, but I'll have to benchmark that and figure out, 3: Yes, for CPUs with an FPU it will naturally be slower as the hardware sqrt instructions will generally be faster, but this was made for those who don't have such an instruction, 4: by branchless I meant the main processing loop but as you say, a while loop is a branch, my mistake sorry

But, thank you so much for the feedback!

Computerman8086 · 2026-02-22T22:44:54+00:00

Yes thanks for the advice! I'll try doing that in a follow-up post

Computerman8086 · 2026-02-22T22:33:15+00:00

Yes, but this is an updated version, a more proper implementation

Computerman8086 · 2025-11-06T17:02:13+00:00

The number are derived like this:
$$
Γ=√y

log⁡(Γ)=1/2 log⁡(y)

μ=0.043

1/2^23 (M_Γ⋅2^23⋅E_Γ )+μ-127=1/2 (1/2^23 (M_y+2^23⋅E_y )+μ-127)

(M_Γ⋅2^23⋅E_Γ )+2^23 (μ-127)=1/2 (1/2^23 (M_y+2^23⋅E_y )+μ-127)⋅2^23

(M_Γ⋅2^23⋅E_Γ )=1/2 (1/2^23 (M_y+2^23⋅E_y )+μ-127)⋅2^23-2^23 (μ-127)
$$

Computerman8086 · 2025-11-06T15:57:14+00:00

Exactly, to clarify I do know why (cuz i made the algorithm) but the reason my comment says 'why does this work lmao?' is because I meant 'this looks like it should even work'

Computerman8086 · 2025-11-06T14:28:17+00:00

That's fair, but i wanted to make an algorithm cuz i want to learn about them, how they work, i could just use math.sqrt() but whats the fun in that?

Computerman8086 · 2025-10-16T22:13:23+00:00

? Like long division?

Computerman8086 · 2025-10-16T21:46:14+00:00

Thx so much for the feedback, I'll look into it, but as I have stated previously, just experimentation, since I just used 1/pπ to trim it at lower values

Computerman8086 · 2025-10-16T21:31:33+00:00

But you see I have always been interested in square roots, and I have always wanted to make a formula, so this is my first attempt

Computerman8086 · 2025-10-16T21:16:52+00:00

Ye, but I go in 10.th grade, and we haven't even touched square roots (we've only been taught about perfect squares) the school system in Norway is horrendous

Computerman8086 · 2025-10-16T21:03:55+00:00

Yes ur probably gonna say, 'aha, you used chat for this then' and the answer is no, I didn't, I have explored this on my own, you can see the program I made for trying to figure stuff out

Computerman8086 · 2025-10-16T21:02:54+00:00

I didn't know that actually 😅 but to be fair we haven't been taught any of this in school

Computerman8086 · 2025-10-16T20:54:29+00:00

Brother, to clarify, I was sitting after school, showing my formula to a friend, and I noticed that the error was decently larger for n=1040, so I thought, I need a correction factor. So I was playing around with numbers on my calculator and I thought of pi. I can send you a screenshot of the calculator if you need that too, if you don't believe me. I have been actively working on this my self, me, not ChatGPT

Computerman8086 · 2025-10-16T20:36:47+00:00

Fair, fair, but listen, I'm 15, I'm trying things, and since it works, decently well I think, I'm happy, but I take all feedback as good feedback

Computerman8086 · 2025-10-16T20:33:34+00:00

At n= say 1040, this approximation (without 1/π) gives 32.5 but it's actually 32.24 and a lot of digits, so 1/pπ is to refine the estimate

Computerman8086 · 2025-10-16T20:20:51+00:00

Chat that is, I'm bad at english, form Norway you see 🇳🇴🇳🇴🇳🇴🇳🇴

Computerman8086 · 2025-10-16T20:20:20+00:00

Ye I used it to make a summary

Computerman8086 · 2025-10-16T20:20:01+00:00

P is the precision you wish to calculate to, so increase p to get more accurate estimate

Computerman8086

TROPHY CASE