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
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!
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?
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
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
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 usingdrop()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 isdroped, 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!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?
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