types - 为什么编译器需要特征的实现才能调用默认的自由函数？

Question

在不采用self的特征上调用默认实现时，为什么需要对实现类型进行注释？
以下是minimal, reproducible example(playground):

mod builder {
    pub trait Builder: Sized {
        fn new() -> Simple {
            Simple
        }
    }

    pub struct Simple;

    impl Builder for Simple {}
}

pub fn main() {
    let _ = builder::Builder::new();
    
    /* Working version */
    // use builder::Builder;
    // let _ = builder::Simple::new();
}

这使:

error[E0283]: type annotations needed
  --> src/main.rs:14:13
   |
3  |         fn new() -> Simple {
   |         ------------------ required by `builder::Builder::new`
...
14 |     let _ = builder::Builder::new();
   |             ^^^^^^^^^^^^^^^^^^^^^ cannot infer type
   |
   = note: cannot satisfy `_: builder::Builder`

error: aborting due to previous error

For more information about this error, try `rustc --explain E0283`.
error: could not compile `playground`.

To learn more, run the command again with --verbose.

E0283的编译器说明未提及默认实现，我同意这是有意义的。但是对于默认实现，为什么需要类型？

Answer 1

Rust中提供的方法与Java中的static方法不同。即使是没有参数的函数和默认实现，实现者也可以覆盖它们。考虑添加另一种实现Builder但覆盖new函数的类型:

struct Noisy {}

impl builder::Builder for Noisy {
    fn new() -> builder::Simple {
        println!("Ahahahah I'm creating a Simple!!!11!");
        builder::Simple
    }
}

fn main() {
    // wait, should this call builder::Simple::new() or Noisy::new()?
    let _ = builder::Builder::new();
}

如果您希望始终具有相同行为的Java static函数的效果，则应使用一个自由函数，正如 prog-fh's answer所建议的那样。