How do I assert an enum is a specific variant if I don't care about its fields?

Question

I\'d like to check enums with fields in tests while ignoring the actual value of the fields for now.

Consider the following example:

enum MyEnum {
           


        

Answer 1

Rust 1.42

You can use std::matches:

assert!(matches!(return_with_fields(), MyEnum::WithFields { .. }));

Previous versions

Your original code can be made to work with a new macro:

macro_rules! is_enum_variant {
    ($v:expr, $p:pat) => (
        if let $p = $v { true } else { false }
    );
}

#[test]
fn example() {
    assert!(is_enum_variant!(return_with_fields(), MyEnum::WithoutFields {..}));
}

Personally, I tend to add methods to my enums:

fn is_with_fields(&self) -> bool {
    match self {
        MyEnum::WithFields { .. } => true,
        _ => false,
    }
}

I also tend to avoid struct-like enums and instead put in extra work:

enum MyEnum {
    WithoutFields,
    WithFields(WithFields),
}

struct WithFields { field: String }

impl MyEnum {
    fn is_with_fields(&self) -> bool {
        match self {
            MyEnum::WithFields(_) => true,
            _ => false,
        }
    }

    fn as_with_fields(&self) -> Option<&WithFields> {
        match self {
            MyEnum::WithFields(x) => Some(x),
            _ => None,
        }
    }
    
    fn into_with_fields(self) -> Option {
        match self {
            MyEnum::WithFields(x) => Some(x),
            _ => None,
        }
    }
}

I hope that some day, enum variants can be made into their own type to avoid this extra struct.