Why is impl Trait on extension methods not allowed? [duplicate]

Is it at all possible to define functions inside of traits as having impl Trait return types? I want to create a trait that can be implemented by multiple structs so that the new() functions of all of them returns an object that they can all be used in the same way without having to write code specific to each one.

trait A {
    fn new() -> impl A;
}

However, I get the following error:

error[E0562]: `impl Trait` not allowed outside of function and inherent method return types
 --> src/lib.rs:2:17
  |
2 |     fn new() -> impl A;
  |                 ^^^^^^

Is this a limitation of the current implementation of impl Trait or am I using it wrong?

As trentcl mentions, you cannot currently place impl Trait in the return position of a trait method.

From RFC 1522:

impl Trait may only be written within the return type of a freestanding or inherent-impl function, not in trait definitions or any non-return type position. They may also not appear in the return type of closure traits or function pointers, unless these are themselves part of a legal return type.

Eventually, we will want to allow the feature to be used within traits [...]

For now, you must use a boxed trait object:

trait A {
    fn new() -> Box<dyn A>;
}

Nightly only

If you wish to use unstable nightly features, you can use existential types (RFC 2071):

// 1.40.0-nightly (2019-11-05 1423bec54cf2db283b61)
#![feature(type_alias_impl_trait)]

trait FromTheFuture {
    type Iter: Iterator<Item = u8>;

    fn example(&self) -> Self::Iter;
}

impl FromTheFuture for u8 {
    type Iter = impl Iterator<Item = u8>;

    fn example(&self) -> Self::Iter {
        std::iter::repeat(*self).take(*self as usize)
    }
}

fn main() {
    for v in 7.example() {
        println!("{}", v);
    }
}

You can get something similar even in the case where it's not returning Self by using an associated type and explicitly naming the return type:

trait B {}
struct C;

impl B for C {}

trait A {
    type FReturn: B;
    fn f() -> Self::FReturn;
}

struct Person;

impl A for Person {
    type FReturn = C;
    fn f() -> C {
        C
    }
}

If you only need to return the specific type for which the trait is currently being implemented, you may be looking for Self.

trait A {
    fn new() -> Self;
}

For example, this will compile:

trait A {
    fn new() -> Self;
}

struct Person;

impl A for Person {
    fn new() -> Person {
        Person
    }
}

Or, a fuller example, demonstrating using the trait:

trait A {
    fn new<S: Into<String>>(name: S) -> Self;
    fn get_name(&self) -> String;
}

struct Person {
    name: String
}

impl A for Person {
    fn new<S: Into<String>>(name: S) -> Person {
        Person { name: name.into() }
    }

    fn get_name(&self) -> String {
        self.name.clone()
    }
}

struct Pet {
    name: String
}

impl A for Pet {
    fn new<S: Into<String>>(name: S) -> Pet {
        Pet { name: name.into() }
    }

    fn get_name(&self) -> String {
        self.name.clone()
    }
}

fn main() {

    let person = Person::new("Simon");
    let pet = Pet::new("Buddy");

    println!("{}'s pets name is {}", get_name(&person), get_name(&pet));
}

fn get_name<T: A>(a: &T) -> String {
    a.get_name()
}

Playground

As a side note.. I have used String here in favor of &str references.. to reduce the need for explicit lifetimes and potentially a loss of focus on the question at hand. I believe it's generally the convention to return a &str reference when borrowing the content and that seems appropriate here.. however I didn't want to distract from the actual example too much.

Why is impl Trait on extension methods not allowed? [duplicate]

Nightly only

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