Derive enum with variants without fields

[azriel91]

EnumIter and EnumString both work by using Default::default() values for any fields on an enum variant. I have a use case where I want to match on the enum's discriminant, but the fields are !Default.

To do this, I propose a #[derive(EnumDiscriminants)] as well as the attribute #[strum_discriminants(derive(..))], which will have the following effect:

struct NonDefault;

#[derive(EnumDiscriminants)]
#[strum_discriminants(
    name(HelloVariants),
    derive(EnumString, EnumIter),
    strum(serialize_all = "snake_case")
)]
enum Hello {
    Variant0,
    Variant1(NonDefault),
    Variant2 { a: NonDefault },
}

// produces:
/// Auto-generated discriminant variants for the `Hello` enum.
#[derive(Clone, Copy, Debug, PartialEq, Eq, EnumString, EnumIter)]
#[strum(serialize_all = "snake_case")]
enum HelloVariants {
    Variant0,
    Variant1,
    Variant2,
}

impl From<Hello> for HelloVariants {
    fn from(val: Hello) -> HelloVariants {
        match val {
            Hello::Variant0 => HelloVariants::Variant0,
            Hello::Variant1(..) => HelloVariants::Variant1,
            Hello::Variant2 { .. } => HelloVariants::Variant2,
        }
    }
}

impl<'_enum> From<&'_enum Hello> for HelloVariants {
    fn from(val: &'_enum Hello) -> HelloVariants {
        match *val {
            Hello::Variant0 => HelloVariants::Variant0,
            Hello::Variant1(..) => HelloVariants::Variant1,
            Hello::Variant2 { .. } => HelloVariants::Variant2,
        }
    }
}

[azriel91]

Done in #34 🎉