WebDec 14, 2024 · This would allow you to call cache.get more than once: fn get (&mut self, buf: &std::vec::Vec) -> Option<&StringObject>. But the returned value will maintain exclusive the borrow of self until dropped. So you wouldn't be able to use the result of the first call after you made the second call. WebAug 12, 2024 · vector of closures: cannot borrow `**h` as mutable, as it is behind a `&` reference ... 1 Cannot borrow value from a hashmap as mutable because it is also borrowed as immutable. 0 Compiler Emitting Message from mutable and immutable reference. 0 Why I can't borrow immutable reference of a mutable reference ...
rust - How to get mutable references to two array elements at …
WebNov 19, 2024 · true_response holds a reference to Response, which means that as long as true_response exists, you cannot do a mutable borrow of Response, which is required by write_response. The issue is basically the same as in the … WebDec 2, 2024 · If your type isn't cloneable, you can transform it into a reference-counted value (such as Rc or Arc) which can then be cloned. You may or may not also need to use interior mutability: struct NonClone; use std::rc::Rc; fn main () { let mut items = vec! incline village mobility committee meeting
rust - Cannot borrow `b` as mutable more than once at a time …
WebNov 30, 2015 · The simplest way to get out from under borrowing problems is to make copies of things, so that you don't need a long-lived borrow; if get_pareto_front_offline returned a Vec< (Vec, (u32, u32))> instead, you wouldn't have this issue. That, or modify to code to not touch neighborhood once you call get_pareto_front_offline. Share. WebMay 6, 2015 · The borrow rules of Rust need to be checked at compilation time, that is why something like mutably borrowing a part of a Vec is a very hard problem to solve (if not impossible), and why it is not possible with Rust. Thus, when you do something like &mut v [i], it will mutably borrow the entire vector. Imagine I did something like WebMar 18, 2024 · 1 Answer Sorted by: 4 After reading up on mutable borrows in for loops it looks like this is the solution: fn place_animal_in_barn (&mut self, animal: Animal<'a>, placement: &str) { for barn in &mut self.barns { if barn.name == placement { barn.animals.push (animal); } } } inbwtss01/ts