CODEWITHSUNDEEP
rust
Axel
Axel

Reputation: 5111

How to move string slice reference to a function so that it's lifetimes dies inside the function passed?

So in the below function I didn't use &content while passing the parameter to func. How is content available after func(content)?

How do you move the content reference to func so that the reference dies inside the func and will not be available afterwards the call if that even makes sense? I want to just free the memory afterwards.

fn main() {
    let content = "Hello World";
    func(content);

    println!("{}", content);
}

fn func(string: &str) {
    println!("{}", &string);
}

Upvotes: 0

Views: 469

Answers (1)

Aplet123
Aplet123

Reputation: 35512

You seem to have a couple misunderstandings in the comments. The fact that you can call func(content) and still use content has nothing to do with the fact that content has a static lifetime. It has to do with the fact that immutable references implement Copy, so they can't be "moved." It's like how this code works despite not having any static references:

fn func<T>(_: T) {}

fn main() {
    let x = 5;
    let y = &x;
    
    func(y);
    println!("{}", y);
}

Even if you have something large that you want to eventually free, you don't need to do anything with the reference. Once the scope is exited (e.g. the function returns), Rust will drop all local variables, freeing any memory if necessary.

Upvotes: 2

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