小编Dav*_*lak的帖子

我想知道是否有任何优雅的解决方案可以获取类似于 Option<&T> 上的 unwrap_or_else 的代码/行为。我的用例是将可选引用传递给函数，如果未使用它，则创建要使用的相同类型的默认值。这是我的代码的精简版本：

#[derive(Debug)]
struct ExpensiveUnclonableThing {}

fn make_the_thing() -> ExpensiveUnclonableThing {
    // making the thing is slow
    // ...
    ExpensiveUnclonableThing {}
}

fn use_the_thing(thing_ref: &ExpensiveUnclonableThing) {
    dbg!(thing_ref);
}

fn use_or_default(thing_ref_opt: Option<&ExpensiveUnclonableThing>) {
    enum MaybeDefaultedRef<'a> {
        Passed(&'a ExpensiveUnclonableThing),
        Defaulted(ExpensiveUnclonableThing),
    }
    let thing_md = match thing_ref_opt {
        Some(thing_ref) => MaybeDefaultedRef::Passed(thing_ref),
        None => MaybeDefaultedRef::Defaulted(make_the_thing()),
    };
    let thing_ref = match &thing_md {
        MaybeDefaultedRef::Passed(thing) => thing,
        MaybeDefaultedRef::Defaulted(thing) => thing,
    };
    use_the_thing(thing_ref);
}

fn use_or_default_nicer(thing_ref_opt: Option<&ExpensiveUnclonableThing>) {
    let thing_ref = thing_ref_opt.unwrap_or_else(|| &make_the_thing());
    use_the_thing(thing_ref); …