CODEWITHSUNDEEP
rust
Tianyi Shi
Tianyi Shi

Reputation: 969

When using generics in Rust, "the size for values of type `str` cannot be known at compilation time"

The following code will work and print Foo("hello"):

#[derive(Debug)]
struct Foo<'a>(&'a str);

impl<'a> From<&'a str> for Foo<'a> {
    fn from(s: &'a str) -> Self {
        Foo(s)
    }
}

fn main() {
    let s: &str = "hello";
    let foo = Foo::from(s);
    println!("{:?}", foo);
}

However, if I use generics, and change all str into T, it won't work:

#[derive(Debug)]
struct Foo<'a, T>(&'a T);

impl<'a, T> From<&'a T> for Foo<'a, T> {
    fn from(s: &'a T) -> Self {
        Foo(s)
    }
}

fn main() {
    let s: &str = "hello";
    let foo = Foo::from(s);
    println!("{:?}", foo);
}

error[E0277]: the size for values of type `str` cannot be known at compilation time
  --> a.rs:12:25
   |
2  | struct Foo<'a, T>(&'a T);
   | ------------------------- required by `Foo`
...
12 |     let foo = Foo::from(s);
   |                         ^ doesn't have a size known at compile-time
   |
   = help: the trait `std::marker::Sized` is not implemented for `str`
   = note: to learn more, visit <https://doc.rust-lang.org/book/ch19-04-advanced-types.html#dynamically-sized-types-and-the-sized-trait>

Isn't &'a T equivalent to &'a str? Why is the compiler seemingly using str instead of &'a str?

Upvotes: 4

Views: 3305

Answers (1)

ekim boran
ekim boran

Reputation: 1819

Type parameters are implicitly Sized in rust. You should add ?Sized constraint explicitly.

#[derive(Debug)]
struct Foo<'a, T: ?Sized>(&'a T);

impl<'a, T: ?Sized> From<&'a T> for Foo<'a, T> {
    fn from(s: &'a T) -> Self {
        Foo(s)
    }
}

Upvotes: 5

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