struct Person<'a> { given: &'a str, sur: &'a mut str, } impl<'a> Person<'a> { fn given(&'a self) -> &'a str { self.given } fn sur_mut(&'a mut self) -> &'a mut str { self.sur } } fn example<'a>(mut person: Person<'a>) { let _one = person.given(); let…

[image] manganhedenbergite: I understand that &'a mut &'a _ should be avoided and is pretty restrictive but, I feel like the problem isn't about that... I take this to mean you understand borrowing forever is bad, but are asking about something else going on in the code. I'm not sure where t…

BTW, you'd never write anything like this in real Rust code [image] Lifetimes in Structs It's a design error. It says that Person doesn't store the name. The name isn't a string, but a string view. The data for the string view is borrowed and the view must …

[image] manganhedenbergite: let _three = person.sur_mut(); is a reborrow of the mutable reference with a shorter lifetime, right? No. It is short form of let r: &'a mut Person<'a> = &mut person; let _three = Person::sur_mut(r); Which fails on first line as the reference constructe…

your code can be extremely simply fixed by doing impl<'a> Person<'a> { fn given<'b>(&'b self) -> &'b str { self.given } fn sur_mut<'b>(&'b mut self) -> &'b mut str { self.sur } } as allows for shorter borrows, where yours only allows for borrows that last for the w…

Thank you for always being there to answer my questions. Your explanations throughout the forum have helped me gain a better understanding of the borrow checker and lifetimes :slight_smile: I have some other questions related to what you’ve written in the rust learning guide, is it okay if I ask th…

[image] manganhedenbergite: I have some other questions related to what you’ve written in the rust learning guide, is it okay if I ask them here? Sure, that's okay.

fn main() { let mut s = String::from("this"); let s_mut = &mut *s; let borrow = &s_mut; f(borrow); // &'a &'b mut str but 'b == 'a => &'b &'b mut str println!("{borrow}"); } fn f<'a: 'b, 'b: 'a>(s: &'a &'b mut str) -> &'b str { *s // &**s } why is this okay?

Nothing need be "borrowed forever" there, due to variance. &'x T is covariant in 'x and T. &'y mut U is covariant in 'y (but invariant in U). So &'x &'y mut str is covariant in 'x and 'y. [1] When you call f(borrow), the argument can coerce to some &'c &'c mut str that needs only be valid for so …

Lifetime question: is my intuition correct?

help

kornel February 16, 2026, 12:21pm 3

BTW, you'd never write anything like this in real Rust code

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Lifetime question: is my intuition correct?

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