rust - 函数重载技巧-编译器无法选择要调用的函数

Question

Rust没有函数重载:您不能使用相同的名称但使用不同的参数来定义多个函数。
但是，我发现了以下技巧，使它看起来像是函数重载。请注意，如何在main()中调用不同类型的example.apply(...):

struct Example;

trait Apply<T, R> {
    fn apply(&self, value: T) -> R;
}

impl Apply<i32, i32> for Example {
    fn apply(&self, value: i32) -> i32 {
        value * 2
    }
}

impl Apply<&str, String> for Example {
    fn apply(&self, value: &str) -> String {
        format!("Hello, {}", value)
    }
}

fn main() {
    let example = Example;

    // Looks like function overloading!
    let one = example.apply(12);
    let two = example.apply("World");

    println!("{}", one);
    println!("{}", two);
}

现在，我想制作Apply特征的两个不同版本:一个用于Copy的值，另一个用于非Copy的值:

struct Example;

struct NotCopy(i32);

// For types which are Copy
trait ApplyCopy<T: Copy, R> {
    fn apply(&self, value: T) -> R;
}

// For types which are not Copy
trait ApplyRef<T, R> {
    fn apply(&self, value: &T) -> R;
}

impl ApplyCopy<i32, i32> for Example {
    fn apply(&self, value: i32) -> i32 {
        value * 2
    }
}

impl ApplyRef<NotCopy, String> for Example {
    fn apply(&self, value: &NotCopy) -> String {
        format!("NotCopy({})", value.0)
    }
}

但是，当我尝试使用它时，出现错误:

fn main() {
    let example = Example;

    let one = example.apply(12);            // Error: multiple `apply` found
    let two = example.apply(&NotCopy(34));  // Error: multiple `apply` found

    println!("{}", one);
    println!("{}", two);
}

错误消息之一:

error[E0034]: multiple applicable items in scope
  --> src/main.rs:30:23
   |
30 |     let one = example.apply(12);            // Error: multiple `apply` found
   |                       ^^^^^ multiple `apply` found
   |
note: candidate #1 is defined in an impl of the trait `ApplyCopy` for the type `Example`
  --> src/main.rs:16:5
   |
16 |     fn apply(&self, value: i32) -> i32 {
   |     ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
note: candidate #2 is defined in an impl of the trait `ApplyRef` for the type `Example`
  --> src/main.rs:22:5
   |
22 |     fn apply(&self, value: &NotCopy) -> String {
   |     ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
help: disambiguate the associated function for candidate #1
   |
30 |     let one = ApplyCopy::apply(&example, 12);            // Error: multiple `apply` found
   |               ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
help: disambiguate the associated function for candidate #2
   |
30 |     let one = ApplyRef::apply(&example, 12);            // Error: multiple `apply` found
   |               ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

编译器提示它不知道要调用哪个apply函数。我希望它能够做出选择:i32不是引用，而是Copy，因此只能调用trait apply的ApplyCopy，并且&NotCopy(34)是引用，并且由于NotCopy不是Copy，所以只能可以调用apply的ApplyRef。
这是Rust编译器的局限性还是我缺少什么？

Answer 1

Is this a limitation of the Rust compiler or am I missing something?

尽管在它们周围有编译器魔术和特殊的大写字母，但从根本上来说，引用仍然是正确的类型，因此任何T都可以是其背后的&Q，并且引用类型可以实现特征。
当NotCopy不是Copy时，是 &NotCopy is Copy 。因此，&NotCopy可以匹配这两种情况。