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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.

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.

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

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.