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Rust generic type that can either own or borrow its content


Rust generic type that can either own or borrow its content

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Rust generic type that can either own or borrow its content
Tag : rust , By : lili
Date : January 11 2021, 05:14 PM

it helps some times std::borrow::Cow
pub enum Cow<'a, B> 
where
    B: 'a + ToOwned + ?Sized, 
 {
    Borrowed(&'a B),
    Owned(<B as ToOwned>::Owned),
}
use std::borrow::Cow;

let mut cow = Cow::Borrowed("foo");
cow.to_mut().make_ascii_uppercase();

assert_eq!(
  cow,
  Cow::Owned(String::from("FOO")) as Cow<str>
);

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generic deserialisation (type-punning) of structs (fighting with the borrow checker)


Tag : rust , By : TBG
Date : March 29 2020, 07:55 AM
wish of those help The argument bs: &[u8] is a slice, and is borrowed. This is a form of temporary ownership, you can't move the data out. *p2 does just that, it moves ownership of that data out.
You need to clone it:
fn from_byte_slice<T: Clone>(bs: &[u8]) -> Option<T> {
    if bs.len() != std::mem::size_of::<T>() {
        None
    } else {
        let p: *const u8 = &bs[0];
        let p2: *const T = p as *const T;
        unsafe {
            Some((*p2).clone())
        }
    }
}

Rust Implement Methods -- Borrow with "getter" method inside mutable borrow method


Tag : development , By : appdelivery
Date : March 29 2020, 07:55 AM
will be helpful for those in need Consider the following example program: , You need to save self.peek() into a temporary
fn write(&mut self) {
    let tmp = self.peek();
    self.output.push(tmp);
}
let tmp1 = &mut self.output;
let tmp2 = self.peek();
Vec<u8>::push(tmp1, tmp2);

How can I define a generic struct in Rust without one of the fields being of the generic type?


Tag : rust , By : Kiltec
Date : March 29 2020, 07:55 AM
may help you . PhantomData can be used "to mark things that "act like" they own a T".
So, you could write:
pub struct Images<T: ImageFormat> {
    path: String,
    phantom: PhantomData<T>, // mark that Image "acts like" it owns a T
}
Images {
    path: ...
    phantom: PhantomData,
}

Rust error: borrow occurs after drop a mutable borrow


Tag : rust , By : Rob
Date : March 29 2020, 07:55 AM
Does that help Your understanding of how scopes and lifetimes work is correct. In Rust Edition 2018, they enabled non-lexical lifetimes by default. Prior to that, the lifetime of inc would have been to the end of the current lexical scope (i.e. the end of the block) even though its value was moved before that.
If you can use Rust version 1.31 or later, then just specify the edition in your Cargo.toml:
[package]
edition = "2018"
rustc --edition 2018 main.rs
#![feature(nll)]
let mut c = 0;
{
    let mut inc = || { c += 1; c };
    drop(inc);
    // scope of inc ends here
}
println!("{}", c);

is there a way to use a generic type alias as the generic type for a function in Rust


Tag : generics , By : Tim
Date : March 29 2020, 07:55 AM
I hope this helps you . Nope since Rust doesn't enforce type bounds on type aliases. Your example is equivalent to this:
type DebuggableFromStr<T> = T;
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