Cannot borrow vector as mutable
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.
Cannot borrow vector as mutable
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WebAug 8, 2024 · Since it is not possible to start the call to IndexMut::index_mut(&mut self, index: Idx) before computing the Idx, there is no reason to start the mutable borrow of v before computing the index. 1 Courtesy of trentcl . WebDec 3, 2024 · The result of *v.index (1) is the value stored at that index, and that value does not require to keep the borrow of v alive. The result of *v.index_mut (1), on the other hand, is a mutable place expression that could theoretically …
WebMay 3, 2016 · To solve this, call tasks.iter_mut () to get an iterator of mutable references. The second problem is calling defining work_one as a method. You already borrow a mutable reference from self when iterating, so you cannot get another borrow. Working example ( playground ): WebSep 25, 2024 · The borrow checker adheres to a set of rules, and the code you posted violates one of them. Here's a direct quote from the Rust book that addresses this exact situation: At any given time, you can have either one mutable reference or any number of immutable references. First you create a mutable variable s1, and borrow it as an …
WebSep 21, 2016 · To change the value that the mutable reference refers to, we have to use the dereference operator ( *) to get to the value in i before we can use the += operator. In addition, you can call the iter_mut method: let mut v = vec! [100, 32, 57]; for i in v.iter_mut () { *i += 50; } See also: WebMar 29, 2024 · This gives me the compiler error: error [E0502]: cannot borrow `vector` as mutable because it is also borrowed as immutable --> src/main.rs:8:9 4 for (i, el) in vector.iter ().enumerate () { ------ immutable borrow occurs here ... 8 vector [i - 1] += el ^^^^^^ mutable borrow occurs here 9 } - immutable borrow ends here
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); } } }
WebNov 19, 2024 · The issue is basically the same as in the following, hopefully simpler example. let mut mutable_string = String::from ("hello"); let immutable_borrow = … earth diver creation storyWebJul 16, 2024 · Your global variables are not mutable. If you want mutable access to those Vec s, you have to wrap them in something that allows that, like Mutex or RwLock. But you should follow @hellow's advice and rethink whether you … earth day eyfsWebFeb 16, 2024 · I understand that this is because borrowing reference to the element also requires borrowing a reference to the vector itself. Therefore, the vector cannot be modified, because that would require borrowing a mutable reference, which is disallowed when another reference to the vector is already borrowed. Here's a simple example earth described in the bibleWebDec 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. dutch brothers real estate departmentWebFeb 8, 2015 · You need to unbox your value before accessing it as a mutable: fn main () { let mut b = Box::new (Vec::new ()); b.push (Vec::new ()); (*b).get_mut (0).unwrap ().push (1); } This is because the . operator uses the Deref trait instead of DerefMut. The best way to achieve this would be: earth destroyerWebSep 16, 2016 · Current Error: error: cannot borrow immutable argument `b` as mutable --> :2:18 1 fn foo (b: &mut u64) { - use `mut b` here to make mutable 2 let x = &mut b; ^ cannot borrow mutably error: aborting due to previous error This error is confusing because: It refers to an argument of type &mut T as "immutable". dutch brothers stock valueWebDec 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! earth day video 2022 for kids