First, I know I can use Box if I want to define a recursive structure. For example,
struct LinkNode {
next: Option<Box<LinkNode>>
}
impl LinkNode{
fn get_next(&self) -> Option<Box<LinkNode>>{
None
}
fn append_next(&mut self, next: LinkNode) -> Self{
self
}
}
But how can I make a trait on these structures via templates or trait object?
Due to the existence of fn append_next(...) -> Self, I cannot create a trait object directly like this:
pub trait Linkable {
fn get_next(&self) -> Option<Box<dyn Linkable>>;
fn append_next(&mut self, next: impl Linkable) -> Self;
}
And we cannot return Option<Box<impl Linkable>> or impl Linkable for fn get_next(&self).
Then I tried the following implementation via generic templates and it does not work.
Because I need to assign the type of T recursively when construct a new LinkNode.
pub trait Linkable<T:Linkable<T> + Clone> : Clone {
fn get_next(&self) -> Option<Box<T>>;
fn append_next(&mut self, next: T) -> Self;
}
I finally implement it in this way, by creating other traits for assistance. And it works well. Again...Is there other better ways?
pub trait Linkable: LinkClone{
fn get_next(&self) -> Option<Box<dyn Linkable>>;
}
pub trait LinkAppend {
fn append_next(&mut self, next: Box<dyn Linkable>) -> Box<dyn Linkable>;
}
pub trait LinkClone{
fn clone_box(&self) -> Box<dyn Linkable>;
}
impl<T> LinkClonefor T
where
T: 'static + Linkable+ LinkAppend + Clone,
{
fn clone_box(&self) -> Box<dyn Linkable> {
Box::new(self.clone())
}
}
impl Clone for Box<dyn Linkable> {
fn clone(&self) -> Box<dyn Linkable> {
self.clone_box()
}
}
BTW, I have some other questions during the exploration above: Why Rust forbids the impl Linkable sugar, like the Box<impl Linkale>? And why returning impl Linkable is forbidden in a trait?
Updated after Ibraheem's answer:
Except the associated type implementation from Ibraheem, it is also fine to work like this. The core idea is to avoid the recursive type declaration in the trait.
pub trait Linkable {
fn get_next<T:Linkable>(&self) -> Next<T>;
fn append_next<T:Linkable>(&mut self, next: Next<T>) -> Self;
}
struct Next<T: Linkable> {
node: T,
}
This is mentioned in another question: Can I define a trait with a type parameter of itself in Rust?
Linkable could have associated type called Next.
pub trait Linkable {
type Next: Linkable;
}
get_next now returns an instance of type Self::Next, and append_next takes Self::Next as a parameter:
pub trait Linkable {
type Next: Linkable;
fn get_next(&self) -> Option<Self::Next>;
fn append_next(&mut self, next: Self::Next) -> &Self;
}
Now you can implement Linkable for Linknode:
impl Linkable for LinkNode {
type Next = LinkNode;
fn get_next(&self) -> Option<Box<LinkNode>> {
None
}
fn append_next(&mut self, next: LinkNode) -> &Self {
self
}
}
Why Rust forbids the impl Linkable sugar, like the Box? And why returning impl Linkable is forbidden in a trait?
You can refer to Is it possible to use impl Trait as a function's return type in a trait definition? for the answer to this question.
Related
Looking for "blanket" implementation of the method(s) for trait.
Let's say for a trait
pub trait A {
fn do_a(&self);
}
want to have boxed method that wraps with box, without introducing any additional traits:
fn boxed(self) -> Box<Self>;
I can have another trait to achieve that (playground)
pub trait A {
fn do_a(&self);
}
pub trait Boxed {
fn boxed(self) -> Box<Self>;
}
impl<T> Boxed for T
where
T: A,
{
fn boxed(self) -> Box<Self> {
Box::new(self)
}
}
However, new trait Boxed is required for that.
You can add boxed directly to A with a default implementation so that structs won't need to implement it themselves:
trait A {
fn do_a(&self);
fn boxed (self) -> Box<Self>
where Self: Sized
{
Box::new (self)
}
}
struct Foo{}
impl A for Foo {
fn do_a (&self) {
todo!();
}
// No need to redefine `boxed` here
}
fn main() {
let foo = Foo{};
let _object: Box<dyn A> = foo.boxed();
}
Playground
How do I make two structures inherit one trait, but with extra parameters?
And I heard that doing so in Rust is not recommended. Why?
struct Foo {name: String,}
struct FooPlus {name: String, lvl: i128,}
trait Trait {
fn show(&self);
fn new(name: &str) -> Self;
// fn new(name: &str, lvl: i128) -> Self;
}
impl Trait for Foo {
fn new(name: &str) -> Self {}
fn show(&self) {}
}
impl Trait for FooPlus {
fn show(&self) {}
fn new(name: &str, lvl: i128) -> Self {}
}
You can make the extra parameter be an associated type to the trait.
trait Trait {
type ConstructParam
fn show(&self);
fn new(name: &str, arg: ConstructParam) -> Self;
}
impl Trait for Foo {
type ConstructParam = ();
fn show(&self) { ... }
fn new(name: &str, arg: ()) -> Self { ... }
}
impl Trait for FooPlus {
type ConstructParam = i128;
fn show(&self) { ... }
fn new(name: &str, arg: i128) -> Self { ... }
}
Note that you have to call the "no-argument" constructor with an extra unit argument, so Foo::new("example name", ()), but that unit argument is zero-sized and thus will almost certainly be optimized to nothing.
Though, as already mentioned in the comments, having new in a trait is odd. How often are you going to polymorphically call new on a type that's generic? If you really need to polymorphically generate instances, that sounds like an application of the factory pattern, not a good use for traits in this way. And if you're not doing it polymorphically, then just make new part of the impl for Foo and FooPlus separately.
I have a function that can handle both owned and borrowed values of some type Foo by accepting a Cow<'_, Foo>. However, I'd like to make it more convenient by allowing to pass in owned or borrowed Foos directly.
Is it possible to have a conversion trait to Cow that's implemented both on a reference type and its owned version?
This is what I tried:
trait Convert<'a> : Clone {
fn into_cow(self) -> Cow<'a, Self>;
}
// Works
impl<'a> Convert<'a> for Foo {
fn into_cow(self) -> Cow<'a, Self> {
println!("owned");
Cow::Owned(self)
}
}
// Doesn't work
impl<'a> Convert<'a> for &'a Foo {
fn into_cow(self) -> Cow<'a, Self> {
println!("borrowed");
Cow::Borrowed(self)
}
}
The borrowed version says that Borrowed expected a &&'a Foo but found a &'a Foo.
Self in the implementation of a trait for &Foo is &Foo, so into_cow() doesn't return the same type in both impls.
One solution would be to change the return type to Cow<'a, Foo>, but that of course limits the trait to only work on Foo.
A better way is to make the owned type a generic parameter of Convert like this:
trait Convert<'a, T: Clone> {
fn into_cow(self) -> Cow<'a, T>;
}
impl<'a, T: Clone> Convert<'a, T> for T {
fn into_cow(self) -> Cow<'a, T> {
println!("owned");
Cow::Owned(self)
}
}
impl<'a, T: Clone> Convert<'a, T> for &'a T {
fn into_cow(self) -> Cow<'a, T> {
println!("borrowed");
Cow::Borrowed(self)
}
}
fn take_foo<'a>(foo: impl Convert<'a, Foo>) {
let cow = foo.into_cow();
}
fn main() {
take_foo(&Foo{});
take_foo(Foo{});
}
Do you ever make use of the ownership, or always reborrow it? If only reborrowing, is std::convert::AsRef what you're looking for?
fn take_foo(foo: impl AsRef<Foo>) {
let foo: &Foo = foo.as_ref();
todo!();
}
TL;DR. Is it possible to automatically implement a trait for an existing type that already has all methods required by the trait?
Long version. Suppose I want to have a generic function that does stack operations on any stack-like type. So I have a trait
pub trait StackLike<T> {
fn is_empty(&self) -> bool;
fn pop(&mut self) -> Option<T>;
fn push(&mut self, value: T);
}
which I can now use like this: pub fn foo(s: &mut dyn StackLike<i32>) { s.push(42); }. So far so good.
There are several existing types that already satisfy my trait, so implementing it is trivial, e.g.
impl<T> StackLike<T> for Vec<T> {
fn is_empty(&self) -> bool { self.is_empty() }
fn pop(&mut self) -> Option<T> { self.pop() }
fn push(&mut self, value: T) { self.push(value) }
}
impl<T, const N: usize> StackLike<T> for SmallVec<[T; N]> {
fn is_empty(&self) -> bool { self.is_empty() }
fn pop(&mut self) -> Option<T> { self.pop() }
fn push(&mut self, value: T) { self.push(value) }
}
This works, but it's a lot of boilerplate. Absolutely nothing interesting happens here: method names, argument types and return type — everything is same.
Question. Is it possible to avoid spelling out all these tautological statements, i.e. do something like the snippet below?
#[implement-trait-automagically]
impl<T> StackLike<T> for Vec<T>;
#[implement-trait-automagically]
impl<T, const N: usize> StackLike<T> for SmallVec<[T; N]>;
Or may be I can mark my trait somehow so that all types that can satisfy it automatically satisfy it (similar to how concepts work in C++)?
Rust doesn't have this kind of "duck typing" but you can put the implementation behind a macro to achieve your stated goals of avoiding all of the boilerplate code while keeping the implementations identical and error-free:
pub trait StackLike<T> {
fn is_empty(&self) -> bool;
fn pop(&mut self) -> Option<T>;
fn push(&mut self, value: T);
}
macro_rules! stacklike_impl {
() => {
fn is_empty(&self) -> bool { self.is_empty() }
fn pop(&mut self) -> Option<T> { self.pop() }
fn push(&mut self, value: T) { self.push(value) }
}
}
impl<T> StackLike<T> for Vec<T> { stacklike_impl!(); }
impl<T, const N: usize> StackLike<T> for SmallVec<[T; N]> { stacklike_impl!(); }
I have a trait, MyGoodTrait, with the function label(&self) -> &str. I want every implementor of MyGoodTrait to also implement Display and FromStr. However, I do not necessarily need Display and FromStr to be supertraits of MyGoodTrait. I would rather somehow have a default implementation of Display and FromStr, which will internally use the label function from MyGoodTrait. That way, every implementor of MyGoodTrait will get Display and FromStr "for free", as if there was a default implementation for those traits.
Here is an example that is similar to what I want to do, but it does not compile:
use std::str::FromStr;
pub trait MyGoodTrait {
fn new() -> Self;
fn label(&self) -> &'static str;
}
impl FromStr for dyn MyGoodTrait {
type Err = String;
fn from_str(s: &str) -> Result<Self, Self::Err> {
Ok(Self::new())
}
}
pub struct A {}
impl MyGoodTrait for A {
fn new() -> Self {
A{}
}
fn label(&self) -> &'static str {
"A"
}
}
pub struct B {}
impl MyGoodTrait for B {
fn new() -> Self {
B{}
}
fn label(&self) -> &'static str {
"B"
}
}
// In this hypothetical, A and B now both have `fmt` and `from_str` functions
Is there a way to write this default implementation of Display and FromStr, such that I do not have to duplicate the code for each struct that implements MyGoodTrait?
Note: My actual use case is that I have a trait which has serde::se::Serialize and serde::de::Deserialize as supertraits. The implementors of my trait will be used as keys in a map, and I will serialize the map to JSON, so I need the implementors to be serialized to Strings. So this may be an example of the XY Problem
TL;DR: You can't.
You can't implement FromStr for dyn SomeTrait because it has a method that returns a Result<Self, _>, hence you can only implement it for types whose size are know at compile time, which is not the case of trait objects.
What you would really want is
impl<T: MyGoodTrait> FromStr for T
But now you might violate the orphan rule. As the compiler explains:
Implementing a foreign trait is only possible if at least one of the types for which is it implemented is local.
Only traits defined in the current crate can be implemented for a type parameter.
But you could do it if FromStr was a local trait instead:
/// Copy of `std::str::FromStr`
trait Foo: Sized {
type Err;
fn from_str(s: &str) -> Result<Self, Self::Err>;
}
impl<T: MyGoodTrait> Foo for T {
type Err = String;
fn from_str(s: &str) -> Result<Self, Self::Err> {
Ok(Self::new())
}
}
Or you could implement it for any specific local type:
impl FromStr for A {
type Err = String;
fn from_str(s: &str) -> Result<Self, Self::Err> {
Ok(Self::new())
}
}