How do I use Option::or with references to Options?

Question

I have the following:

fn foo(f: &Option<Huge>) {}

fn bar(a: &Option<Huge>, b: &Option<Huge>) {
    foo(a.or(b));
}

Huge is some big struct that I don't want to copy or clone. This does not work because .or() takes a and b by value.

Is there an easy solution? I can probably do something like this:

foo(if a.is_some() { a } else { b });

Surely there is a better way?

Answer 1

&Option<T>几乎从来都不是一个有用的类型，但它可以转换为Option<&T>usingas_ref。

这意味着以下代码有效：

fn foo(f: Option<&Huge>) {}

fn bar(a: Option<&Huge>, b: Option<&Huge>) {
    foo(a.or(b));
}