Is it possible to bind all the struct members of an enum variant to a single variable?

user_5164339 0
Viewed 545

Let say I define and instantiate an enum as follows:

enum MyEnum {
    EmptyVariant,
    TupleVariant(u8),
    StructVariant {
        key: u8,
        value: char,
    }
}

let instance = MyEnum::StructVariant{key: 8, value: 'a'};

Is it possible to match against this variant without destructuring? For example, instead of doing:

if let MyEnum::StructVariant{key, value} = instance {
    eprintln!("key, value = {}, {}", key, value);
}

I would rather write something like:

if let MyEnum::StructVariant{VARIANT_MEMBERS} = instance {
    eprintln!("key, value = {}, {}", VARIANT_MEMBERS.key, VARIANT_MEMBERS.value);
}

In this example, writing out the members of the struct variant is benign, but in the case where the variant has many members it makes the code difficult to read.

rust enums syntax
1 Answers

I don't think it is possible as of today.

I did see the following pattern, which in practice achieves what you asked for just with an extra intermediary type:

Instead place all the desired members of the StructVariant in an actual struct type, and have your enum use that struct as the only field of the StructVariant.

struct MyVariant {
    key: u8,
    value: char
}

enum MyEnum {
    EmptyVariant,
    TupleVariant(u8),
    StructVariant(MyVariant)
}

if let MyEnum::StructVariant(x) = instance {
    eprintln!("key, value = {}, {}", x.key, x.value);
}

This also proves handy when you want to pass the members of StructVariant around, to other functions or types.

user_2485340
Related