Is there any way to implement a trait with the same struct of different generic constraints in Rust?

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Suppose I have a struct S<T> and I want implement it to trait Default with different constraints of generic type T:

struct S<T> { ... }


impl<T> Default for S<T> 
where
    T: Ord
{
    fn default() -> Self {
        Self::new()
    }
}

impl<T> Default for S<T>
where
    T: Ord + Default
{
    fn default() -> Self {
        Self::build(T::default())
    }
}

However, the compiler complains:

error[E0119]: conflicting implementations of trait std::default::Default for type S<_>

Is there any way to compile such code?

rust traits
1 Answers

There is no way to do this in stable Rust.

You can use the (unstable and unsound!) specialization feature on nightly Rust:

#![feature(specialization)]

struct S<T> {
    x: Option<T>
}

impl<T> S<T> {
    fn new() -> Self { Self { x: None } }
    fn build(x: T) -> Self { Self { x: Some(x) } }
}


impl<T> Default for S<T> 
where
    T: Ord
{
    default fn default() -> Self {
        Self::new()
    }
}

impl<T> Default for S<T>
where
    T: Ord + Default
{
    fn default() -> Self {
        Self::build(T::default())
    }
}

https://play.rust-lang.org/?version=nightly&mode=debug&edition=2018&gist=ee24c383878f7a32c53d12211b69ab4f

user_89220
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