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The parts contain borrowed strings that point back into the line. Therefore Rust ensures that line lives at least as long as the parts. However, in the second version the result of unwrap() is a temporary, and the temporary lives only until the end of the statement. Rust does not automatically extend the lifetime of temporaries for you (that.
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error[E0597]: `stream` does not live long enough --> src/main.rs:47:35 | 47 | match handle_request(&mut stream){ | ^^^^^ borrowed value does not live long enough. 54 | } | - borrowed value only lives until here | note: borrowed value must be valid for the anonymous lifetime #1 defined on the function body at 43:1.
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When I compile the code below, rustc complains: error[E0597]: `vec` does not live long enough --> longenough.rs:14:19 | 14 | let vec_ref = vec.iter_mut().map(|b| &mut **b).collect(); | ^^^ borrowed value does not live long enough 15 | f(vec_ref); 16 | } | - `vec` dropped here while still borrowed | = note: values in a scope are dropped in the opposite order they are created However I don't.
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This isn't exactly a limitation. It's simply how static type safety is supposed to work. It's exactly the same as how you wouldn't expect to be able to assign 42 to a parameter declared to be a String and then expect the compiler to deduce that it should actually have a numeric type. You declared it to be a parameter i.e. have a lifetime that outlives the function.
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I've read over the chapters on references, borrowing, and lifetimes a couple times, but some things are still a little slow to click for me. I'm receiving a borrowed value does not live long enough error, and figured I could find some help here in the community. Here is my module, string.rs: extern crate co.
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Where 's is "chosen" by the caller of foo. Then you try to: b.do_sth(&temp); // where `temp` is a local you don't return. Since the caller defines 's, they might give you an arbitrary long lifetime. You need to be able to choose an arbitrary short lifetime. So need something that implements the trait for any lifetime.
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@JaredLanders Chapter 10.3 is about lifetime syntax and talks about 'static.Here's one relevant excerpt: Lifetime annotations don't change how long any of the references live.They] describe the relationships of the lifetimes of multiple references to each other without affecting the lifetimes.
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These word references cannot outlive the if let block, which means you cannot use them as keys in the hash map. The easiest fix is to have words_matches own its keys. Without borrows there are no lifetimes, and thus no lifetime issues. let mut words_matches: HashMap
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An important note for anyone wanting to contribute & improve this diagnostic: The rules around default trait object lifetimes are way more intricate (see the Rust Reference) than just `dyn View` without an explicit lifetime implies `'static`, not `'_` (as found in the PR description). In general, the default depends on a lot of factors. @rustbot label A-lifetimes D-terse
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There are a few ways to make a borrowed value expire. One way is to assign the borrowed value to another variable. For example, the following code creates a mutable borrowed value of the variable `x` and assigns it to the variable `y`: rust. let x = 5; let mut y = &x; // y is now a mutable borrowed value of x.
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You should make the item type in the channel String instead of &'static str.The static string slice type can only be used with immutable global constants such as string literals.
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rust-lang / rust Public. Notifications Fork 12k; Star 92.8k. Code; Issues 5k+ Pull requests 615; Actions; Projects 2; Security; Insights New issue Have a question about this project?. borrowed value does not live long enough in closure #76627. Open 95th opened this issue Sep 12, 2020 · 4 comments Open
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A lifetime means "how long the variable lives". You only need to think about lifetimes with references. This is because references can't live longer than the object they come from. For example, this function does not work: let my_string = String ::from( "I am a string" ); &my_string // ⚠️. fn main () {}
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This won't work if the lifetime is 'static for presumably obvious reasons -- but again, the only thing you do know about the lifetime is that it's longer than the body of your function, so in fact, no lifetime at all can work with this pattern so long as the caller is choosing the lifetimes.
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