Generic enum

Define generic enums like Option<T>.

Generic enums

The two enums you use most are generic:

enum Option<T> {
    Some(T),
    None,
}

enum Result<T, E> {
    Ok(T),
    Err(E),
}

You can write your own generic enum too. For example, to express a search result independent of the type being searched for:

enum Sonuc<T> {
    Bulundu(T),
    Yok,
}

Methods on a generic enum

Methods on a generic enum are also written with impl<T>; the trait bound decides what can be done with the content:

use std::fmt::Display;

impl<T: Display> Sonuc<T> {
    fn goster(&self) -> String {
        match self {
            Sonuc::Bulundu(v) => format!("bulundu: {v}"),
            Sonuc::Yok => String::from("yok"),
        }
    }
}

Exercise Search result

Write the Sonuc<T> enum above and the goster method. Print the results for Sonuc::Bulundu(42) and Sonuc::<&str>::Yok separated by | .

Show example solution

use std::fmt::Display;

enum Sonuc<T> {
    Bulundu(T),
    Yok,
}

impl<T: Display> Sonuc<T> {
    fn goster(&self) -> String {
        match self {
            Sonuc::Bulundu(v) => format!("bulundu: {v}"),
            Sonuc::Yok => String::from("yok"),
        }
    }
}

fn main() {
    let a = Sonuc::Bulundu(42);
    let b = Sonuc::<&str>::Yok;
    println!("{} | {}", a.goster(), b.goster());
}

Test yourself

Why doesn't <code>let s = Sonuc::Yok;</code> compile on its own?
- Enums can't be generic
- It compiles
- T can't be inferred; it must be given like Sonuc::<i32>::Yok
- Yok is an invalid variant
Show answer

T can't be inferred; it must be given like Sonuc::<i32>::Yok — The Yok variant carries no information about T; the type is either written explicitly or inferred from later use.

Specify the type

let s = Sonuc::____::Yok;

Show answer

let s = Sonuc::<i32>::Yok;

Value and type

Some(5)

Option<i32>
Ok::<u8, String>(1)

Result<u8, String>
Sonuc::Bulundu("a")

Sonuc<&str>
Sonuc::<f64>::Yok

Sonuc<f64>

Next lesson: Turbofish

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