I extracted this example from my actual issue:
fn main() {
let vec = vec![2u8, 4, 8];
let vec_ = vec.as_slice();
let y = Box::new(*vec_);
println!("{:?}", y);
}
My intention is to transform &[u8] to [u8]. I tried to dereference it by * but I got error:
error[E0277]: the size for values of type `[u8]` cannot be known at compilation time
--> src/main.rs:4:22
|
4 | let y = Box::new(*vec_);
| ^^^^^ doesn't have a size known at compile-time
|
= help: the trait `std::marker::Sized` is not implemented for `[u8]`
= note: to learn more, visit <https://doc.rust-lang.org/book/ch19-04-advanced-types.html#dynamically-sized-types-and-the-sized-trait>
= note: required by `std::boxed::Box::<T>::new`
error[E0277]: the size for values of type `[u8]` cannot be known at compilation time
--> src/main.rs:4:13
|
4 | let y = Box::new(*vec_);
| ^^^^^^^^ doesn't have a size known at compile-time
|
= help: the trait `std::marker::Sized` is not implemented for `[u8]`
= note: to learn more, visit <https://doc.rust-lang.org/book/ch19-04-advanced-types.html#dynamically-sized-types-and-the-sized-trait>
= note: all function arguments must have a statically known size
= help: unsized locals are gated as an unstable feature
Although the compiler give noted link. I read it carefully but cant get useful information to help me to solve my question: how to transform a &[u8] to [u8]?
