this post was submitted on 31 May 2024
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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.

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[–] stuckgum@lemmy.ml 43 points 5 months ago (70 children)

Yet another security issue that Rust would solve.

[–] DacoTaco@lemmy.world 5 points 5 months ago (14 children)

Serious question, how would using rust avoid this? Rust still has reference types in the background, right? Still has a way to put stuff on the heap too? Those are the only 2 requirements for reusing memory bugs

[–] sleep_deprived@lemmy.world 30 points 5 months ago (13 children)

This is a use-after-free, which should be impossible in safe Rust due to the borrow checker. The only way for this to happen would be incorrect unsafe code (still possible, but dramatically reduced code surface to worry about) or a compiler bug. To allocate heap space in safe Rust, you have to use types provided by the language like Box, Rc, Vec, etc. To free that space (in Rust terminology, dropping it by using drop() or letting it go out of scope) you must be the owner of it and there may be current borrows (i.e. no references may exist). Once the variable is droped, the variable is dead so accessing it is a compiler error, and the compiler/std handles freeing the memory.

There's some extra semantics to some of that but that's pretty much it. These kind of memory bugs are basically Rust's raison d'etre - it's been carefully designed to make most memory bugs impossible without using unsafe. If you'd like more information I'd be happy to provide!

[–] paysrenttobirds@sh.itjust.works -1 points 5 months ago (2 children)

The way I understand it, it is a bug in C implementation of free() that causes it to do something weird when you call it twice on the same memory. Maybe In Rust you can never call free twice, so you would never come across this bug. But, also Rust probably doesn't have the same bug.

My point is it seems it is a bug in the underlying implementation of free(), not to be caught by the compiler, and can't Rust have such errors no matter its superior design?

[–] Feyd@programming.dev 10 points 5 months ago (1 children)

The way that rust attempts to prevent this class of error is not by making an implementation of free that is safe to call twice, but by making the compiler refuse to compile programs where free could be called twice on a pointer.

Anyway, use after free doesn't depend on a double free. It just means that the program frees memory but keeps the pointer (which now points at memory that could contain unrelated data at some future point in time) and if someone trying to exploit the program finds a way to induce the program to read or write to that memory they may be able to access data they are not expected to, or write data to be used by a different part of the program that they shouldn't be able to

[–] paysrenttobirds@sh.itjust.works 2 points 5 months ago

Thanks, I understand the problem with using memory after it's been freed and possibly access it changed by another part of the process. I guess I was confused by the double free explanation I read, which didn't really say how it could be exploited, but I think you are right it still needs to be accessed later by the original program, which would not happen in Rust.

[–] Nibodhika@lemmy.world 9 points 5 months ago (1 children)

Not really, the issue is that C/C++ is not memory safe, i.e. it allows you to access memory that has already been freed. Consider the following C++ code:

int* wrong() {
  int data  = 10;
  return &data;
}

If you try to use it it looks correct:

int* ptr = wrong();
std::cout << *ptr << std::endl;

That will print 10, but the memory where data was defined has been freed, and is no longer in control of the program. Meaning that if something else allocated that memory they can control what my program does.

Consider that on that example above later in the program we do:

user.access_level = *ptr;

If someone manages to get control of that memory between when we freed it and we used it they can make the access_level of the user be whatever they want.

This is a problem with C/C++ allowing you to access memory that has been freed, which is why C/C++ programmers need to be extra careful.

[–] paysrenttobirds@sh.itjust.works 3 points 5 months ago

Thank you, that is very clear.

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