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!
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?
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.
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 'a annotation for strA and the return value, and 'b for 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 :
DataBaseUser variable1 = GetDataBaseUser(20);
userService.Users.Add(variable1);
variable1 = null; // or free?
[end of function scope here, reference to heap now in list ]
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!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?
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.
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 :