What #[derive(IsVariant)] generates

When an enum is decorated with #[derive(IsVariant)], for each variant foo in the enum, a public instance method is_foo(&self) -> bool is generated. If you don't want the is_foo method generated for a variant you can put the #[is_variant(ignore)] attribute on that variant.

Example usage

# use derive_more::IsVariant;
#
#[derive(IsVariant)]
enum Maybe<T> {
    Just(T),
    Nothing
}

assert!(Maybe::<()>::Nothing.is_nothing());
assert!(!Maybe::<()>::Nothing.is_just());

What is generated?

The derive in the above example generates code like this:

# enum Maybe<T> {
#     Just(T),
#     Nothing
# }
impl<T> Maybe<T>{
    #[must_use]
    pub const fn is_just(&self) -> bool {
        matches!(self, Self::Just(..))
    }
    #[must_use]
    pub const fn is_nothing(&self) -> bool {
        matches!(self, Self::Nothing)
    }
}