this post was submitted on 31 May 2024
284 points (98.0% liked)
Linux
48224 readers
111 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
Thanks for the response. Ive heard of rust's compiler being very smart and checking a ton of stuff. Its good thing it does, but i feel like there are things that can cause this issues rust cant catch. Cant put my finger on it.
What would rust do if you have a class A create something on the heap, and it passes this variable ( by ref ? ) to class B, which saves the value into a private variable in class B. Class A gets out of scope, and would be cleaned up. What it put on the heap would be cleaned up, but class B still has a reference(?) to the value on the heap, no? How would rust handle such a case?
You use lifetimes to annotate parameters and return values in order to tell the compiler about how long things must last for your function to be valid. You can link a specific input with the output, or explicitly separate them. If you don't give lifetimes the language uses some basic rules to do it for you. If it can't, eg it's ambiguous, then it's a compile error and you need to do it manually.
It's one of the harder concepts of rust to explain succinctly. But imagine you had a function that took strA and strB, used strB to find a subsection of strA, and then return a slice of strA. That slice is tied to strA. You would use
'aannotation for strA and the return value, and'bfor strB.Rust compiler will detect the lifetime being shorter than expected.
Also, ownership semantics. Think c++ move semantics. Only one person is left with a good value, the previous owners just have garbage data they can't use anymore. If you created a thing on the heap and then gave it away, you wouldn't have it anymore to free at the end. If you want to have "multiple owners" then you need ref counting and such, which also stops this problem of premature freeing.
Edit: one more thing: reference rules. You can have many read-only references to a thing, or one mutable reference. Unless you're doing crazy things, the compiler simply won't let you have references to a thing, and then via one of those references free that thing, thereby invalidating the other references.
Thats interresting, thanks! Stuff for me to look into!
I also think halfway through the conversation i might have given the impression i was talking about pointers, while it was not my intention to do so. That said, the readonly/mutable reference thing is very interresting!
Ill look into what rust does/has that is like the following psuedocode :
No problem. I'm no guru and I'm currently on Zig but I think learning some Rust is a really fast way to hone skills that are implied by other languages.
It's not like C where you have control over when you can make references to data. The compiler will stop you from making references in the cases where a memory bug would be possible.
Rust simply doesn't allow you to have references to data that goes out of scope (unless previously mentioned hoops are jumped through such as an explicitly declared unsafe block). It's checked at compile time. You will never be able to compile the program.
Rust isn't C. Rust isn't C++. The memory-safe-ness of it is also not magic, it's a series of checks in the compiler.
That sounds odd. That also means that a mapper, command, service,... can never return a class object or entity. Most of the programming world is based on oop o.O
Keep in mind im not talking about the usage of pointers, but reference typed variables.
Oh sure, I'm still learning so I thought you meant references as in pointers like in C++. But also, Rust isn't a strictly object oriented language either. It shares a lot of similar features, but they aren't all the typical way you'd do things in an OOP language. You should check out the chapter of the Rust book for ownership.