this post was submitted on 11 Apr 2024
162 points (95.0% liked)
Linux
51569 readers
479 users here now
From Wikipedia, the free encyclopedia
Linux is a family of open source Unix-like operating systems based on the Linux kernel, an operating system kernel first released on September 17, 1991 by Linus Torvalds. Linux is typically packaged in a Linux distribution (or distro for short).
Distributions include the Linux kernel and supporting system software and libraries, many of which are provided by the GNU Project. Many Linux distributions use the word "Linux" in their name, but the Free Software Foundation uses the name GNU/Linux to emphasize the importance of GNU software, causing some controversy.
Rules
- Posts must be relevant to operating systems running the Linux kernel. GNU/Linux or otherwise.
- No misinformation
- No NSFW content
- No hate speech, bigotry, etc
Related Communities
Community icon by Alpár-Etele Méder, licensed under CC BY 3.0
founded 5 years ago
MODERATORS
you are viewing a single comment's thread
view the rest of the comments
view the rest of the comments
The more the code is used, the faster it ought to be. A function for an OS kernel shouldn't be written in Python, but a calculator doesn't need to be written in assembly, that kind of thing.
I can't really speak for Rust myself but to explain the comment, the performance gains of a language closer to assembly can be worth the headache of dealing with unsafe and harder to debug languages.
Linux, for instance, uses some assembly for the parts of it that need to be blazing fast. Confirming assembly code as bug-free, no leaks, all that, is just worth the performance sometimes.
But yeah I dunno in what cases rust is faster than C/C++.
C/C++ isn’t really faster than Rust. That’s the attraction of Rust; safety AND speed.
Of course it also depends on the job.
https://benchmarksgame-team.pages.debian.net/benchmarksgame/box-plot-summary-charts.html
You're talking about two languages, one is C, the other is C++. C++ is not a superset of C.
Yes thank you. But my statement remains true nevertheless.