当枚举变体已知时解包内部类型

Question

I have this enum type:

enum Animal {
    Dog(i32),
    Cat(u8),
}

Now I have a function that takes this type as parameter. I know (for some reason) that the input is always a Cat . I want to achieve this:

fn count_legs_of_cat(animal: Animal) -> u8 {
    if let Animal::Cat(c) = animal { c } else { unreachable!() }
}

Can I write this shorter and/or more idiomatic?

Answer 1

Not really. What I have seen is introducing a new struct for each enum variant, and then methods on the enum to decompose it:

struct Dog(i32);
struct Cat(u8);

enum Animal {
    Dog(Dog),
    Cat(Cat),
}

impl Animal {
    fn cat(self) -> Cat {
        if let Animal::Cat(c) = self { c } else { panic!("Not a cat") }
    }

    fn dog(self) -> Dog {
        if let Animal::Dog(d) = self { d } else { panic!("Not a dog") }
    }
}

// Or better an impl on `Cat` ?
fn count_legs_of_cat(c: Cat) -> u8 {
    c.0
}

Of course, you don't need the struct and you could just return the u8 , but that may get hard to track.

There's a glimmer of better support for this in the future, however. I think it's the "efficient code reuse" RFC , but better described in the blog post Virtual Structs Part 3: Bringing Enums and Structs Together . The proposal would be to allow Animal::Cat to be a standalone type, thus your method could accept an Animal::Cat and not have to worry about it.

---

Personally, I almost always prefer to write the infallible code in my inherent implementation and force the caller to panic:

impl Animal {
    fn cat(self) -> Option<Cat> {
        if let Animal::Cat(c) = self {
            Some(c)
        } else {
            None
        }
    }

    fn dog(self) -> Option<Dog> {
        if let Animal::Dog(d) = self {
            Some(d)
        } else {
            None
        }
    }
}

And I'd probably use a match

impl Animal {
    fn cat(self) -> Option<Cat> {
        match self {
            Animal::Cat(c) => Some(c),
            _ => None,
        }
    }

    fn dog(self) -> Option<Dog> {
        match self {
            Animal::Dog(d) => Some(d),
            _ => None,
        }
    }
}

Answer 2

I found one single macro is the best way to solve the problem (in recent Rust).

Macro Definition

    macro_rules! cast {
        ($target: expr, $pat: path) => {
            {
                if let $pat(a) = $target { // #1
                    a
                } else {
                    panic!(
                        "mismatch variant when cast to {}", 
                        stringify!($pat)); // #2
                }
            }
        };
    }

Macro Usage

let cat = cast!(animal, Animal::Cat);

Explanation:

• #1 The if let exploits recent Rust compiler's smart pattern matching. Contrary to other solutions like into_variant and friends, this one macro covers all ownership usage like self , &self and &mut self . On the other hand {into,as,as_mut}_{variant} solution usually needs 3 * N method definitions where N is the number of variants.

• #2 If the variant and value mismatch, the macro will simply panic and report the expected pattern.

• The macro, however, does not handle nested pattern like Some(Animal(cat)) . But it is good enough for common usage.

Answer 3

试试enum-as-inner crate，它完全符合 Shepmaster 的回答。

Answer 4

I wrote a small macro for extracting the known enum variant:

#[macro_export]
macro_rules! extract_enum_value {
  ($value:expr, $pattern:pat => $extracted_value:expr) => {
    match $value {
      $pattern => $extracted_value,
      _ => panic!("Pattern doesn't match!"),
    }
  };
}

let cat = extract_enum_value!(animal, Animal::Cat(c) => c);

However, I'm not sure if this fits into your need.